Physics I Homework Page

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This page has all of the homework for the first semester of College Physics. You can go directly to each section using the links above or by using the menu.

Kinematics

    No Acceleration

  1. How much time will it take for a car to travel 1000 m at a speed of 25 m/sec?
  2. How far will a car with a speed of 10 m/sec travel in 60 sec?
  3. What is the speed of a car that travels 150 m in 30 sec?
  4. Here is a teaser problem from Click and Clack the Tappet Brothers (Car Talk on NPR).

    ROAD TRIP TRIGONOMETRY (week of 8/23/14, Link)

    RAY: Last Fall, Tommy and I decided that we were going to take a trip north to see the foliage. Tommy drove the first 40 miles. I drove the rest of the way. We looked at the foliage for three or four minutes, then decided to head home. We took the same route home. On the way back, Tommy drove the first leg of the trip and I drove the last 50 miles. I got home and my wife said, "Who did the driving?" I explained that Tommy drove the first 40 miles, then I drove the rest of the way. On the way back, Tommy drove the first leg of the trip, and I drove the last 50 miles. She said, "But who did most of the driving?" I told her, "You can figure it out. In fact, you can even figure out how much more of the driving was done by that person." And that's the question. Who drove the most -- and how many more miles did that person drive?

    Horizontal Acceleration (Friction)

  5. A car traveling on dry concrete at 8.94 m/sec (about 20 mi/hr) sees a cat in the road, applies the brakes, and slides to a stop next to the cat. (a) How long did it take for the car to reach the cat after the car started sliding? (b) How far did the car slide? As stated in the text, a car sliding on dry concrete has an acceleration of -6.86 m/secē (from experiment).
    Answer

    v = 0 m/sec   vo = 8.94 m/sec   s = ?   a = -6.86 m/sec²   t = ?

    v = vo + at;   t = (v - vo)/a = (0 + 8.94 m/sec)/(6.86 m/sec²) = 1.3 sec

    v² = vo² + 2as;  

    s = [v² - vo²]/2a = [0 - (8.94 m/sec)²]/(2)(-6.86 m/sec²) = 5.8 m

    Numerical Answers
    1.3 sec 5.8 m
  6. If a car left skid marks that were 30 m long on a flat wet asphalt road. (a) How fast was the car moving before the brakes were applied? (b) How long did the car slide? A typical car sliding on wet asphalt has an acceleration of -3.43 m/secē (this number is from experiment).
    Answer

    v = 0 m/sec   vo = ? m/sec   s = 30 m   a = -3.43 m/sec²   t = ?

    v² = vo² + 2as;

    vo = [v² - 2as]½ = [0 + (2)(3.43 m/sec²)(30 m)]½ = 14.35 m/sec

    v = vo + at;   t = (v - vo)/a = (0 + 14.35 m/sec)/(3.43 m/sec²) = 4.18 sec

    Numerical Answers
    14.35 m/sec 4.18 sec
  7. A wooden box was on a conveyer belt that moved the box at a constant horizontal speed. The end of the conveyer belt connected to a level floor where the box would slide and come to rest. The workers have 2.1 seconds from the time the box hits the floor to the time when it stops. Wood on this floor has an acceleration of -4.86 m/secē. (a) What was the velocity of the wooden box when it first hit the floor? (b) How far did it slide?
    Answer

    v = 0 m/sec   vo = ? m/sec   s = ? m   a = -4.86 m/sec²   t = 2.1 sec

    v = vo + at;   vo = v - at = 0 + (4.86 m/sec²)(2.1 sec) = 10.2 m/sec

    v² = vo² + 2as;  

    s = [v² - vo²]/2a = [0 - (10.2 m/sec)²]/(2)(-4.86 m/sec²) = 10.7 m

    Numerical Answers
    10.2 m/sec 10.7 m
  8. Vertical Acceleration (Gravity)

  9. A ball was freely falling for 2.0 seconds and had a velocity of 20.0 m/sec at the end of that time. (a) What was the initial velocity of the ball? (b) What distance did the ball travel during that 2.0 sec?
    Answer

    v = -20.0 m/sec   vo = ? m/sec   s = ? m   a = -9.8 m/sec²   t = 2.0 sec

    vo = v - at = (-20.0 m/sec) + (9.8 m/sec²)(2.0 sec) = -0.4 m/sec

    v² = vo² + 2as;

    s = [v² - vo²]/2a = [(-20 m/sec)² - (-0.4 m/sec)²]/(2)(-9.8 m/sec²) = -20.4 m

    Numerical Answers
    -0.4 m/sec -20.4 m
  10. Uncle John has a pitching machine that will throw balls at specified velocities. Being somewhat curious, he wondered how fast a ball would have to be propelled to just hit the ceiling 10 meters above his head. He didn't want it to hit the machine when it came back down, so he wondered how much time he would have to move the machine. He knew you, taking physics and all, could figure this out. What do you tell him?
    Answer

    From start to max:

    v = 0 m/sec   vo = ? m/sec   s = 10 m   a = -9.8 m/sec²   t = ? sec

    v² = vo² + 2as; vo = [v² - 2as]½ = [0 - (2)(-9.8 m/sec²)(10 m)]½ = 14 m/sec

    v = vo + at;   t = (v - vo)/a = (0 - 14 m/sec)/(-9.8 m/sec²) = 1.43 sec

    Total time is the time to go to max plus the time to come back down:

    2(1.43 sec) = 2.86 sec

    Numerical Answers
    14 m/sec 2.86 sec
  11. While walking with curious Aunt Mary, you came across an ancient well that had long since gone dry. Aunt Mary wanted to know how deep the well was. She turned to you, the physics expert, to tell her how to figure out this mystery. You told her to use the timer on her watch to clock a rock from when it was released (from rest) until it hit the bottom. She said it took 4 seconds. How deep was the well?
    Answer

    v = ? m/sec   vo = 0 m/sec   s = ? m   a = -9.8 m/sec²   t = 4.0 sec

    s = vot + ½at² = 0 + ½(-9.8 m/sec²)(4 sec)² = -78.4 m

    The well is 78.4 m deep.

    Also:

    v = vo + at = (0 m/sec) - (9.8 m/sec²)(4.0 sec) = -39.2 m/sec

    Numerical Answers
    78.4 m
  12. Not taking anything for granted, Uncle John wanted to test the method used for calculating the depth of the well in the last problem. He took his pitching machine and pointed it directly down from a very tall building. One time he gave the ball zero velocity and found that it took 3 seconds to reach the ground. The next time he gave the ball a speed of 3.1 m/s and found that it took 2.7 seconds. How high was the building calculated to be each time? Does the mathematical model work?
    Answer

    1st Time: v = ? m/sec   vo = 0 m/sec   s = ? m   a = -9.8 m/sec²   t = 3.0 sec

    s = vot + ½at² = 0 + ½(-9.8 m/sec²)(3 sec)² = -44.1 m

    2nd Time: v = ? m/sec   vo = -3.1 m/sec   s = ? m   a = -9.8 m/sec²   t = 2.7 sec

    s = vot + ½at² = (-3.1 m/sec)(2.7 sec) + ½(-9.8 m/sec²)(2.7 sec)² = -44.1 m

    Both calculations say the building is 44.1 m high. The model works!

    Numerical Answers
    44.1 m
  13. How long does it take a freely falling object to reach a velocity of -50 m/s if it starts at rest? How far did it drop?
    Answer

    v = -50 m/sec   vo = 0 m/sec   s = ? m   a = -9.8 m/sec²   t = ? sec

    t = (v - vo)/a = [(-50 m/sec) - 0]/(-9.8 m/sec²) = 5.1 sec

    s = vot + ½at² = 0 + ½(-9.8 m/sec²)(5.1 sec)² = -127.5 m

    It dropped 127.5 m.

    Numerical Answers
    5.1 sec 127.5 m
  14. Horizontal and Vertical Acceleration (Projectiles)

    Here are some mathematical relationships that will be helpful for some of these two dimensional problems.

  15. A frog jumps off of a 0.5 m bank at 25o above horizontal with a speed of 3 m/sec. (a) Determine the distance from the bank to where the frog hits the water. (b) How long was the frog in the air?
    Answer

    Calculate components of the initial velocity in x and y.

    In y: voy = vosin25° = (3 m/sec)(sin25°) = 1.27 m/sec

    In x: vox = vocos25° = (3 m/sec)(cos25°) = 2.72 m/sec

    yx
    v? m/sec2.72 m/sec
    vo1.27 m/sec2.72 m/sec
    s-0.5 m? m
    a-9.8 m/sec0 m/sec
    t? sec? sec

    Make a table from jump to water.

    Solve for v in the y-direction, then solve for t in the y-direction. Use the time from the y-direction in the x-direction to determine the distance in the x-direction.

    vy = [voy² + 2aysy]½ = [(1.27 m/sec)² + (2)(-9.8 m/sec²)(-0.5 m)]½

    = -3.38 m/sec

    t = (vy - voy)/ay = [(-3.38 m/sec) - 1.27 m/sec]/(-9.8 m/sec²) = 0.474 sec

    sx = voxt + ½at² = (2.72 m/sec)(0.474 sec) + 0 = 1.29 m

    Numerical Answers
    1.29 m 0.474 sec
  16. An airplane traveling horizontally 1000 m above the ground, drops a CARE package 2000 m horizontally from a village which caused it to land right in the middle of the village. (a) What was the horizontal speed of the airplane? (b) What was the magnitude and direction of the final velocity?
    Answer
    yx
    v? m/sec? m/sec
    vo0 m/sec? m/sec
    s-1000 m2000 m
    a-9.8 m/sec0 m/sec
    t? sec? sec

    Make a table from drop to ground.

    Solve for v in the y-direction, then solve for t in the y-direction. Use the time from the y-direction in the x-direction to determine the velocity in the x-direction.

    vy = [voy² + 2aysy]½ = [0 + (2)(-9.8 m/sec²)(-1000 m)]½

    = -140 m/sec

    t = (vy - voy)/ay = [(-140 m/sec) - 0]/(-9.8 m/sec²) = 14.3 sec

    vox = sx/t = (2000 m)/(14.3 sec) = 140 m/sec

    For the final velocity:

    v = [(vx)² + (vy)²]½ = [(-140 m/sec)² + (140 m/sec)²]½ = 198 m/sec

    θ = tan-1[(vx)/vy)] = tan-1(140/140) = 45°

    v = 198 m/sec at an angle of 45° below horizontal.

    Numerical Answers
    140 m/sec 198 m/sec at 45° below horizontal.
  17. A projectile is fired with an initial velocity that has a horizontal component of 160 m/sec and a vertical component of 75 m/sec. (a) Determine the initial velocity (magnitude and direction). (b) What is the maximum height of the projectile? (c) What is the range of the projectile assuming it starts and ends at the same vertical height.
    Answer

    Get the initial velocity from the initial components.

    v0 = [(vox)² + (voy)²]½ = [(160 m/sec)² + (75 m/sec)²]½ = 176.7 m/sec

    θ = tan-1[(voy)/vox)] = tan-1(75/160) = 25.1°

    The initial velocity is 176.7 m/sec at 25.1° above horizontal.

    yx
    v0 m/sec160 m/sec
    vo75 m/sec160 m/sec
    s? m? m
    a-9.8 m/sec0 m/sec
    t? sec? sec

    Make a table going from start to max.

    Solve for s in the y-direction to get the maximum height.

    sy = [(vy)² - (voy)²)]/[2ay] = [(0 - (75 m/sec)²)]/[2(-9.8 m/sec²)] = 287 m

    t = (vy - voy)/ay = [(0 m/sec) - 75 m/sec]/(-9.8 m/sec²) = 7.65 sec (to max)

    The total time is 2(7.65 sec) = 15.3 sec

    For the distance in x (the range): sx = vot = (160 m/sec)(15.3 sec) = 2449 m

    Numerical Answers
    The initial velocity is 176.7 m/sec at 25.1° above horizontal.

    Distance to Max = 287 m, Range = 2449 m

  18. When a spring gun is held at an angle of 40o above the horizontal, the range of the projectile is 30 m. (a) What is the initial velocity of the projectile? (b) What is the maximum height? (c) How long was the projectile in the air? Assume the projectile starts and ends at the same height.
    Answer (b and c)

    From part (a):

    vox = vocosθ m/sec = (17.3 m/sec)cos40° = 13.3 m/sec

    voy = vosinθ = (17.3 m/sec)sin40° = 11.12 m/sec

    Make a table going from start to max.

    yx
    v0 m/secvx = vox m/sec
    vo11.12 m/sec13.3 m/sec
    s? m15 m
    a-9.8 m/sec0 m/sec
    tt sect sec

    sy = [(vy)² - (voy)²)]/[2ay] = [(0 - (11.2 m/sec)²)]/[2(-9.8 m/sec²)]

    = 6.4 m

    t = (vy - voy)/ay = [(0 m/sec) - 11.2 m/sec]/(-9.8 m/sec²) = 1.14 sec (to max)

    The total time is 2(1.14 sec) = 2.28 sec

    Answer (a)
    yx
    vvsinθ m/secvcosθ m/sec
    vovosinθ m/secvocosθ m/sec
    s0 m30 m
    a-9.8 m/sec0 m/sec
    tt sect sec

    Make a table going from start to end.

    This will require solving two equations and two unknowns. Solve for t in both the x-direction and the y-direction, then set those expressions equal to each other.

    In x: 30 m = (vosinθ m/sec)(t) or t = (30 m)/(vosinθ m/sec)
    In y: sy = voyt + ½ayt² = [voy + ½ayt]t

    0 = [voy + ½ayt]t says that either t = 0 or [voy + ½ayt] = 0

    Time equal to zero is before it starts, so the second expression is needed to agree with the physical situation. So, from y:

    t = -voy/½a = -(vosinθ)/½a

    That means:

    -(vosinθ)/½a = (30 m)/(vosinθ m/sec) or -(vo²)(sinθ)(cosθ) = ½a(30)

    vo = -[½a(30)]/[(sinθ)(cosθ)]½ = [(15 m)(9.8 m/sec²)]/[(sin40°)(cos40°)]½

    = 17.3 m/sec

    Numerical Answers
    17.3 m/sec 6.4 m 2.28 sec
  19. An object is thrown from a bridge at an angle of 40o from the horizontal to a person that is 10 m from the base of the bridge. The bridge is 15 m above the ground. (a) How long will the object be in the air? (c) What will the final velocity be (magnitude and direction)?
    Answer

    Make a table going from start to end.

    yx
    vvsinθ m/secvx = vox m/sec
    vovosinθ m/secvocosθ m/sec
    s-15 m10 m
    a-9.8 m/sec0 m/sec
    t? sec? sec

    Solve for the time in x and substitute the resulting expression into the distance expression for y.

    In x:   t = (10 m)/(vocos40°)

    sy = (voy)t + ½at²

    Substiute t from above and then solve for the initial velocity, vo.

    -15 m = (vo)(sin40°)[(10 m)/(vocos40°)] + ½(-9.8 m/sec²)[(10 m)/(vocos40°)]²

    vo = 5.975 m/sec; voy = 3.84 m/sec; vox = 4.58 m/sec;

    t = (10 m)/(4.58 m/sec) = 2.18 sec

    vy = 3.84 m/sec - (9.8 m/sec²)(2.18 sec) = -17.52 m/sec

    v = [(4.58 m/sec)² + (-17.52 m/sec)²]½ = 18.1 m/sec

    θ = tan-1(-17.52/4.58) = 75.3°

    Numerical Answers
    t = 2.18 sec, v = -17.52 m/sec at 75.3° below the positive x-axis.
  20. An archer shoots an arrow at an angle of 10o above horizontal. The arrow stays in the air for 2.1 sec before hitting a target that is at the same height as where the arrow is released. (a) What are the initial and final velocities (magnitude and direction)? (b)How far away from the archer is the target?
    Answer
    yx
    v0 m/secvx = vox m/sec
    vo? m/sec? m/sec
    s? m? m
    a-9.8 m/sec0 m/sec
    t1.05 sec1.05 sec

    Symmetry makes this problem much easier.

    Make a table going from start to max.

    0 = voy + at;   voy = -at = (9.8 m/sec²)(1.05 sec) = 10.29 m/sec

    vo = (voy)/sinθ = (10.29 m/sec)/sin10° = 59.3 m/sec at 10°

    vox = (vo)(cosθ) = (59.3 m/sec)(cos10°) = 58.4 m/sec

    Because of the symmetry of this problem the final velocity in y will be the negative of the initial velocity in y. The velocity in x remains constant. θ = tan-1(-10.29/58.4) = -10°

    v = [(vx)² + (vy)²]½ = 59.3 m/sec at -10°

    sx = (58.4 m/sec)(2.1 sec) = 123 m

    Numerical Answers
    59.3 m/sec at 10° 59.3 m/sec at -10° 123 m
  21. EXTRA PROBLEMS

  22. Among Aunt Mary's interests is rocketry. She built a rocket that was propelled by pressure from a chemical reaction. On a calm day she decided to launch the rocket, desiring to obtain a maximum height. Just as she was about to release the rocket, her young son came by, knocked the pad from vertical to some unknown angle, and caused the rocket to be launched. Her timing device functioned properly letting her know that the rocket was in the air for 8.0 seconds. She found the rocket at a horizontal distance of 60 meters. (a) What angle was the initial velocity? [79.2° above horizontal] (b) What was the maximun height of the rocket? [78.4 m] (c) What was the final velocity of the rocket just before hitting the ground? [39.9 m/sec at 79.2° below horizontal]
  23. A wooden box was on a conveyer belt that moved the box at a constant horizontal speed. The end of the conveyer belt connected to a level stone floor where the box would slide and come to rest. The workers have 5 seconds from the time the box hits the floor to the time when it stops. Wood on stone has an acceleration of -3.92 m/secē. What was the velocity of the wooden box when it first hit the stone floor?
  24. Calculate the distance that is required to stop a bus that has its brakes applied while going 20 m/s and accelerates at -4 m/sec².
  25. A policeman investigating the scene of an accident measured skid marks made by a bus that are 3.5 m long. Assume a bus with its brakes locked decelerates at 4 m/s². Calculate how fast the bus was initially traveling. How long did an observer have to react before the bus stopped?
  26. A coin is dropped from rest. It was observed by a person 5 meters from the ground to take 1.3 seconds before it hits the ground. How high was the person who dropped the coin? How long was the coin in the air?
  27. Two people on a very high balcony wanted to settle an argument by flipping a coin. The coin was tossed up 2 meters from the edge of the balcony, but not caught and it fell toward the ground. As luck would have it, a physics class was testing a new radar device that was sensitive enough to determine the velocity of the coin to be 15 m/s at a height of 20 m from the ground. They calculated the height that the coin must have fallen from, but it was not the same height as a balcony or a window. What height was the balcony, and why were the students confused?
    Answer
  28. An object is thrown off of a 15 m cliff at an angle of 30° above horizontal. It hits the ground after 10 seconds. Calculate the initial velocity, the distance it travels horizontally, and the maximum height above the cliff. [voy = 34 m/sec, vo = 68 m/sec, sx = 588.89 m]
  29. On the freeway, how much distance should you leave between your car and the one ahead if you speed is 60 mph and you can decelerate at 10 m/s²? Is this reasonable? What might be wrong with this calculation?

Static and Dynamic Equilibrium Homework

  1. A 3.6 kg flower pot hangs from a tree suspended by a rope. What is the tension in the rope?
    Answer

    Force up = Force down

    T = mg = (3.6 kg)(9.8 m/sec²) = 35.28 N

    Numerical Answers
    35.28 N
  2. Refer to the diagram below for the next few problems.

  3. If θ1 = θ2 = 25° and F = 5.6 N, what are the tensions T1 and T2?
    Answer

    T1x = T2x   and   T1y + T2y = F

    T1cosθ1 = T2cosθ2   and   T1sinθ1 + T2sinθ2 = F

    T2 = (T1cosθ1)/cosθ2   and   T1sinθ1 + [(T1cosθ1)/cosθ2]sinθ2 = F

    T1[sinθ1 + [(cosθ1)(tanθ2)] = F     T1 = F/[sinθ1 + [(cosθ1)(tanθ2)]

    T1 = (5.6 N)/[sin25° + [(cos25°)(tan25°)] = 6.63 N

    T2 = (T1cosθ1)/cosθ2 = (6.63 N)(cos25°)/(cos25°) = 6.63 N

    Numerical Answers
    6.63 N 6.63 N
  4. If the tensions T1 and T2 are both equal to 38 N and θ1 = θ2 = 15°, what is the force F?
    Answer

    T1x = T2x   and   T1y + T2y = F

    F = T1y + T2y = T1sinθ1 + T2sinθ2

    F = (38 N)(sin15°) + (38 N)(sin15°) = 19.67 N

    Numerical Answers
    19.67 N
  5. If F = 12 N and the tensions T1 and T2 are both equal to 22 N, what are the angles θ1 and θ2?
    Answer

    T1x = T2x   and   T1y + T2y = F

    T1cosθ1 = T2cosθ2   and   T1sinθ1 + T2sinθ2 = F

    T1cosθ1 = T2cosθ2   and   T1 = T2 cosθ1 = cosθ2 θ1 = θ2

    Tsinθ1 + Tsinθ1 = F sinθ1 = F/(2T)

    θ1 = sin-1[F/2T] = sin-1[(12 N)/{2(22 N)}] = 15.8° = θ2

    Numerical Answers
    15.8° 15.8°
  6. If θ1 = 20°, θ2 = 30°, T1 = 8 N, and T2 = 5 N, what is the force F?
    Answer

    F = T1y + T2y = T1sinθ1 + T2sinθ2

    F = (8 N)(sin20°) + (5 N)(sin30°) = 5.24 N

    Numerical Answers
    5.24 N
  7. If F = 40 N, θ2 = 56.31°, and T2 = 30 N, what are T1 and θ1?
    Answer

    T1x = T2x   and   F = T1y + T2y

    T1cosθ1 = T2cosθ2   and   F = T1sinθ1 + T2sinθ2

    T1 = T2cosθ2/cosθ1   and   F = (T2cosθ2/cosθ1)sinθ1 + T2sinθ2

    F = T2cosθ2tanθ1 + T2sinθ2     tanθ1 = (F - T2sinθ2)/T2cosθ2

      tanθ1 = [(40 N) - (30 N)sin56.31°]/[(30 N)cos56.31°] = 0.9037

    θ1 = tan-10.9037 = 42.1°

    T1 = [F - T2sinθ2]/sinθ1 = [(40 N) - (30 N)sin56.31°]/sin42.1° = 22.43 N

    Numerical Answers
    42.1° 22.43 N
  8. A 600 N gymnast, in a display of upper body and hand strength, brought both rings together and held them in one hand. At that point there were 30° between the ropes (assume them to be the same length). What tension is in each rope?
    Answer

    T1x = T2x   and   F = T1y + T2y

    T1cosθ1 = T2cosθ2   and   F = T1sinθ1 + T2sinθ2

    T1 = T2 = T   and   θ1 = θ2 = θ = 90° - 15° = 75°

    F = Tsinθ + Tsinθ   = 2Tsinθ

      T = F/(2sinθ) = (600 N)/(2sin75°) = 311 N

    Numerical Answers
    311 N
  9. One at a time, members of a scout troop cross a river by moving hand over hand on a horizontal rope attached to trees on opposite sides of the river that are 16 m apart. What will the tension in the rope be on each side of a 500 N scout at a horizontal distance of 4 m if the rope is pulled down 1 m? (Answer: 1505 N and 1546 N)
    Answer

    Using the distances:

    tanθ1 = (1 m)/(4 m) θ1 = 14°   and   tanθ2 = (1 m)/(12 m) θ2 = 4.76°

    T1x = T2x   and   F = T1y + T2y

    T1cosθ1 = T2cosθ2   and   F = T1sinθ1 + T2sinθ2

    T1 = T2cosθ2/cosθ1   and   F = (T2cosθ2/cosθ1)sinθ1 + T2sinθ2

    F = T2cosθ2tanθ1 + T2sinθ2     T2 = F/[cosθ2tanθ1 + sinθ2]

      T2 = (500 N)/[cos4.76°tan14° + sin4.76°] = 1509 N

      T1 = T2cosθ2/cosθ1 = T1 = (1509 N)cos4.76°/cos14° = 1549 N

    Numerical Answers
    1509 N 1549 N
  10. A 35 N block slides down a 25° ramp at a constant speed of 1.0 m/sec. How much force is keeping the block from accelerating?
    Answer

    F = (mg)sinθ = (35 N)sin25° = 14.8 N

    Numerical Answers
    14.8 N

Net Force Problems

  1. A 150 g mass and a 400 g mass are hanging from the ends of a string that is over a friction-free pulley. What is the acceleration of the masses and what is the tension in the string?
    Answer

    Whole System:

    netForce   = Fg2 - Fg1 = ma

    = (0.4 kg)(9.8 m/sec²) - (0.15 kg)(9.8 m/sec²) = 2.45 N

    = 2.54 N = ma

    a = (2.45 N)/(0.55 kg) = 4.45 m/sec²

    Isolate the 0.4 kg Mass:

    netForce   = Fg2 - T2 = ma

    T2   = Fg2 - ma = (0.4 kg)(9.8 m/sec²) - (0.4 kg)(4.45 m/sec²)

    T2   = 2.14 N

    Numerical Answers
    4.45 m/sec² 2.14 N
  2. A 3 kg box is placed on a friction-free horizontal table and connected by a string over a friction-free pulley to a 2 kg box that is hanging off of the edge of the table. What is the acceleration of this system and what is the tension in the string?
    Answer

    Whole System:

    netForce   = Fg2 = (m1 + m2)a

    = (2 kg)(9.8 m/sec²) = 19.6 N = (5 kg)a

    a = (19.6 N)/(5 kg) = 3.92 m/sec²

    Isolate the 2 kg Mass:

    netForce   = Fg2 - T = m2a

    T   = Fg2 - m2a = (2 kg)(9.8 m/sec²) - (2 kg)(3.92 m/sec²)

    T   = 11.76 N

    Numerical Answers
    3.92 m/sec² 11.76 N
  3. A 20 N block accelerates at 2.34 m/sec² down a 30° ramp. How much friction is slowing down the block? What is the coefficient of friction in this case?
    Answer

    Fgx = Fgsinθ ;   Fgy = Fgcosθ ;   Fn = Fgy ;   fk = μkFn

    m = (20 N)/(9.8 m/sec²) = 2.04 kg

    x-axis (along the incline):

    netForce   = Fgx - fk = ma

    fk = Fgx - ma = (20 N)sin30° - (2.04 kg)(2.34 m/sec²) = 5.22 N

    y-axis (perpendicular to the plane):

    fk = μkFn

    μk = fk/Fn = (5.22 N)/[(20 N)cos30°]

    μk = 0.3

    Numerical Answers
    5.22 N 0.3
  4. A 5 kg mass is on a 25° ramp and has a coefficient of kinetic friction of 0.11 with the ramp. A string is pulling on this 5 kg mass due to a second mass that is connected to the other end of the string and hanging over the end of the ramp (using a friction-free pulley). If the second, hanging mass is 12 kg, what is the acceleration and what is the tension in the string?
    Answer

    F12 = (12 kg)(9.8 m/sec²) = 117.6 N ;   F5 = (5 kg)(9.8 m/sec²) = 49 N

    F5x = F5sinθ = (49 N)sin25° = 20.71 N ; F5y = F5cosθ = (49 N)cos25° = 44.41 N

    Fn = F5y = 44.41 N ;   fk = μkFn = (0.11)(44.41 N) = 4.88 N

    Whole System:

    netForce   = F12 - fk - F5x = (m12 + m5)a

    = 117.6 N - 4.88 N - 20.71 N = 92 N = (17 kg)a

    a = (92 N)/(17 kg) = 5.41 m/sec²

    Isolate the 12 kg Mass:

    netForce   = F12 - T = m12a

    T   = F12 - m12a = 117.6 N - (12 kg)(5.41 m/sec²) = 52.68 N

    Numerical Answers
    5.41 m/sec² 52.68 N

Centripetal Force Problems

  1. A car of mass 1.5 x 10³ kg rounds a circular turn of radius 20 m. If the road is flat and the coefficient of friction is 0.50 between the tires and the road, how fast can the car travel without skidding?
    Answer

    fs = Fc ;   fs = μsFn ;   Fc = mv²/r ; Fn = mg

      μsmg = mv²/r   μsg = v²/r

      v = [rμsg]½ = [(0.5)(9.8 m/sec²)(20 m)]½ = 9.9 m/sec

    Numerical Answers
    9.9 m/sec²
  2. A race track designed for average speeds of 240 km/h is to have a turn with a radius of 975 m. To what angle must the track be banked so that cars traveling 240 km/h have no tendency to slip sideways?
    Answer

    See example #2 in the Centripetal Force section of Newton's Laws for a diagram.

    Fgx = Fcx

    Fgsinθ = Fccosθ

      tanθ = Fc/Fg = (m)(v²/r)/mg = v²/rg

    = [(240 km/h)(1000 m/km)(1 h/3600 sec)]² /(9.8 m/sec²)(975 m)

    = .466

      θ = tan-1(.466) = 24.9°

    Numerical Answers
    24.9°
  3. What is the maximum speed of a 1500 kg car at the top of a hill that has a radius of curvature of 30 m?
    Answer

    At the top of the hill the gravitational force pulling the car down must at least equal the centripetal force that would make the car go straight (at a tangent to the curve, away from the center). The maximum speed would be when those two forces are equal: Fg = Fc.

      mg = mv²/r

      v =[rg]½ = [(30 m)(9.8 m/sec²)]½ = 17.15 m/sec²

    Numerical Answers
    17.15 m/sec²
  4. EXTRA PROBLEMS
  5. A 20 N box sits on a 40 degree inclined plane and is connected to a rope that goes over a pulley at the top end of the inclined plane and is then connected to a 2.9 kg weight that hangs freely. (a) How fast will the system accelerate and (b) what will the tension in the rope be?
  6. Two masses, M1 = 2 kg and M2 = 3 kg, are suspended on opposite ends of a rope that has been placed over a pulley. What is the acceleration and what is the tension in the rope?
  7. Calculate the acceleration when a 20 N sled is being pulled over smooth ice with a horizontal force of 3 N.
  8. Find the tension in the rope and the acceleration when a 6 g mass and an 8 g mass are suspended from opposite ends of a rope that is placed over a pulley.
  9. A 60 N box is placed on a 35 degree inclined plane and connected to a rope that provides a constant upward force of 10 N. What is the acceleration of the box if the coefficient of kinetic friction is 0.3?
  10. A 5 kg mass is accelerated up a 35 degree inclined plane by the tension in a rope that goes over a pulley and is connected to a second 3 kg mass that hangs freely. If the system accelerates at 0.14 m/sec², what is the coefficient of kinetic friction between the 5 kg mass and the inclined plane? What is the tension in the rope?
    Answer
  11. If F = 40 N, T1 = 20 N, and T2 = 30 N, what are the angles θ1 and θ2?
  12. What is the maximum speed you can go around a slippery curve with a circular radius of 40 m if the curve is banked at 10°?
    Answer
  13. What is the maximum speed you can go around a curve with a circular radius of 40 m if the curve is banked at 10° and there is friction with μs = 0.7?
    Answer
    With and Without Friction Answers
  14. What is the minimum speed you can go around a curve with a circular radius of 40 m if the curve is banked at 10° and there is friction with μs = 0.7? What if μs is only 0.15?
    Answer

Work and Conservation Laws

    Work Problems

  1. How much work is done when a 0.75 kg book is raised 1.5m?
    Answer

    W = Fs = mgh = (0.75 kg)(9.8 m/sec²)(1.5 m) = 11.03 J

    Numerical Answers
    11.03 J
  2. How much work is done when a 1.5 kg box slides down a 20° friction-free incline a distance of 2 m?
    Answer

    W = Fs = Fxsx = [(1.5 kg)(9.8 m/sec²)(sin20°)](2 m) = 10.06 J

    Numerical Answers
    10.06 J
  3. What distance was traveled if a 25 N force required 18 J of work to move the object?
    Answer

    W = Fs s = W/F = (18 J)/(25 N) = 0.72 m

    Numerical Answers
    0.72 m
  4. Conservation of Energy Problems

  5. What is the final velocity when sliding down a 20° incline if: vi = 7 m/sec, s = 2 m, and W = 0?
    Answer

    KEi + PEi = KEf + PEf + W     ½mvi² + mghi = ½mvf² + mghf + Fs

    Set h = 0 at the bottom of the incline.

    ½m(7 m/sec)² + m(9.8 m/sec²)[(2 m)(sin20°)] = ½mvf² + 0 + 0

    The mass cancels out

    vf² = 2{½(7 m/sec)² + (9.8 m/sec²)[(2 m)(sin20°)]}

    vf = (62.41 m²/sec²)½ = 7.9 m/sec

    Numerical Answers
    7.9 m/sec
  6. A ball starting from rest rolls down a friction-free "roller-coaster" track starting at a height of 12 m and goes through a dip at a height of 3 m and then goes back up on top of a "hill" at a height of 8 m. What is the speed at the top of the 8 m hill?
    Answer

    KEi + PEi = KEf + PEf + W     ½mvi² + mghi = ½mvf² + mghf + Fs

    Set h = 0 at 8 m. Friction-free W = 0. From rest vi = 0.

    0 + m(9.8 m/sec²)(4 m) = ½mvf² + 0 + 0

    The mass cancels out

    vf = [2{9.8 m/sec²)(4 m)]½ = 8.85 m/sec

    Numerical Answers
    8.85 m/sec
  7. What is the final velocity when a 3.3 kg mass slides down a 50° incline if: vi = 0, s = 3.3 m, µk = 0.12?
    Answer

    KEi + PEi = KEf + PEf + W     ½mvi² + mghi = ½mvf² + mghf + W

    Set h = 0 at the bottom of the incline. Fg = mg

    W = fk(s) = µkFn(s) = µkFgy(s) = µk(mgcosθ)(s)

    With this W the mass will still cancel out. hi = (3.3 m)(sin50°) = 2.528 m.

    0 + (9.8 m/sec²)(2.528 m) = ½vf² + 0 + (0.12)(9.8 m/sec²)(cos50°)(3.3 m)

    0 + 24.774 m²/sec² = ½vf² + 0 + 2.495 m²/sec²

    vf = [2(24.774 m²/sec² - 2.495 m²/sec²)]½ = 6.68 m/sec

    Answer with Picture
  8. A slide for young children is 1.75 m high at an angle of 40° from the horizontal with an additional horizontal length of slide at a height of 0.25 m from the ground. (a) If the slide is friction-free, what is the velocity of the child at the bottom of the incline, starting from rest? (b) If the coefficient of friction between the child and the horizontal part of the slide is 0.11, how long should the horizontal section be to stop the child completely?
    (b)

    Part (b): Start: After the incline, End: where it stops on the horizontal section.

    KEi + PEi = KEf + PEf + W     ½mvi² + mghi = ½mvf² + mghf + W

    Set h = 0 at the bottom of the incline. hi = hf = 0.   vi = 5.42 m/sec.   vf = 0.

    W = Fs = fks = Fnµks = Fgµks = (mg)(µk)(s)

    ½mvi² + 0 = 0 + 0 + (mg)(µk)(s)

    Divide both sides by the mass and multiply both sides by 2.

    vi² + 0 = 0 + 0 + (2)(g)(µk)(s)

    s = (vi²)/[(2)(g)(µk)] = (5.42 m/sec)²/[(2)(9.8 m/sec²)(0.11)] = 13.6 m

    Answers: (a)

    Part (a):

    KEi + PEi = KEf + PEf + W     ½mvi² + mghi = ½mvf² + mghf + Fs

    Set h = 0 at the bottom of the incline. hi = 1.75 m - 0.25 m = 1.5 m. W = 0.

    Fg = mg   Fgx = mgsinθ

    With W = 0 the mass will cancel out.

    0 + (9.8 m/sec²)(1.5 m) = ½vf² + 0 + 0

    vf = [2(9.8 m/sec²)(1.5 m)]½ = 5.42 m/sec

    Answer with Picture
  9. What will be the final velocity of a rock that is thrown off of a 15 m cliff with a speed of 20 m/sec at an angle of 30° above the horizontal? Use conservation of energy.
    Answer

    KEi + PEi = KEf + PEf + W     ½mvi² + mghi = ½mvf² + mghf + Fs

    Set h = 0 at the bottom. Friction-free (ignore air friction) W = 0.

    vix = (20 m/sec)(cos30°) = 17.32 m/sec   viy = (20 m/sec)(sin30°) = 10 m/sec

    The mass again cancels out. In the vertical direction:

    ½(10 m/sec)² + (9.8 m/sec²)(15 m) = ½vfy² + 0 + 0

    vfy = {2[(50 m²/sec²) + (147 m²/sec²)]}½ = 19.8 m/sec

    vf = [vfx² + vfy²]½ = [(17.32 m/sec)² + (19.8 m/sec)²]½ = 26.3 m/sec

    Numerical Answers
    26.3 m/sec
  10. A car runs out of gas just as it goes over the top of a hill. If the car is traveling at 75 km/hr at the top of the hill and the hill is 20 m high: (a) Will the car make it to the top of the next hill, which is 25 m high? (b) What is the minimum speed that the car can have and still make it to the top of the next hill?
    Answer

    KEi + PEi = KEf + PEf + W     ½mvi² + mghi = ½mvf² + mghf + Fs

    Set h = 0 at 20 m. hf = 5 m. Assume friction-free W = 0.

    vi = (75 km/hr)(1000 m/1 km)(1 hr/3600 sec) = 20.83 m/sec  

    The mass again cancels out. The velocity at the top of the second hill must be at least zero (a). The minimum speed is when the final velocity is zero (b).

    ½(20.83 m/sec)² = ½vf² + (9.8 m/sec²)(5 m) + 0

    vf = [2{(217 m²/sec²) - (49 m²/sec²)}]½ = 18.33 m/sec

    The car will go over the top of the hill at 18.33 m/sec. (Part a)

    Part b:   vf = 0     vi = [2ghf]½ = 9.9 m/sec

    Numerical Answers
    The car will go over the top of the hill at 18.33 m/sec. 9.9 m/sec
  11. A 500 N boy slides down a 30° incline starting from rest. His final velocity is 1.0 m/sec after going 5.0 m down the incline. What is the coefficient of friction between the boy and the incline?
    Answer

    KEi + PEi = KEf + PEf + W     ½mvi² + mghi = ½mvf² + mghf + Fs

    Set h = 0 at 5 m down the ramp. hi = (5 m)sin30° = 2.5 m. hf = 0 m.

    vi = 0.   vf = 1 m/sec.   m = Fg/g = (500 N)/(9.8 m/sec²) = 51 kg

    W = Fs = (Fgx - fk)s = (Fgsinθ - Fnµk)s = (Fgsinθ - Fgyµk)s

    = (Fgsinθ - [Fgcosθ]µk)s = Fg[sinθ]s - Fg[cosθ]µks

    = (500 N)(sin30°)(5 m) - (500 N)(cos30°)(µk)(5 m) = 1250 J - (2165 J)µk

    0 + mghi = ½mvf² + 0 + W

    0 + (51 kg)(9.8 m/sec²)(2.5 m) = ½(51 kg)(1 m/sec)² + 0 + 1250 J - (2165 J)µk

    µk = [½(51 kg)(1 m/sec)² + 1250 J - (51 kg)(9.8 m/sec²)(2.5 m)]/(2165 J) = 0.012

    Numerical Answers
    0.012
  12. Conservation of Momentum Problems

  13. A 400 g air track car and a 350 g air track car are fitted with velcro so that they remain together upon collision. If the 400 g car is traveling at 0.12 m/sec and collides with the 350 g car when it is at rest, what will be the final velocity of the two cars (which remain together)?
    Answer

    pi = pf     p1i + p2i = p1f + p2f     m1v1i + m2v2i = m1v1f + m2v2f

    In this case: m1v1i + m2v2i = (m1 + m2)vf

    vf = (m1v1i + m2v2i)/(m1 + m2) = [(0.4 kg)(0.12 m/sec) + 0]/(0.4 kg + 0.35 kg)

    vf = 0.064 m/sec

    Numerical Answers
    0.064 m/sec
  14. A 400 g air track car and a 350 g air track car are fitted with velcro so that they remain together upon collision. If the 400 g car is traveling at 0.12 m/sec and collides with the 350 g car when it is traveling toward the 400 g car at 0.06 m/sec, what will be the final velocity of the two cars (which remain together)?
    Answer

    pi = pf     p1i + p2i = p1f + p2f     m1v1i + m2v2i = m1v1f + m2v2f

    In this case: m1v1i + m2v2i = (m1 + m2)vf

    vf = (m1v1i + m2v2i)/(m1 + m2)

    vf = [(0.4 kg)(0.12 m/sec) + (0.35 kg)(-0.06 m/sec)]/(0.4 kg + 0.35 kg)

    vf = 0.036 m/sec

    Numerical Answers
    0.036 m/sec
  15. A 55 kg boy and a 45 kg girl face each other on ice skates. The girl pushes the boy, who moves away at a speed of 2 m/sec. What is the girl's speed?
    Answer

    pi = pf     p1i + p2i = p1f + p2f     m1v1i + m2v2i = m1v1f + m2v2f

    In this case: 0 = m1v1f + m2v2f

    v2f = -m1v1f/m2

    v2f = -[(55 kg)(2 m/sec)]/(45 kg) = -2.44 m/sec

    v2f = 2.44 m/sec away from the boy

    Numerical Answers
    -2.44 m/sec
  16. A 50 kg girl throws a 1.5 kg physics book horizontally with a speed of 1.0 m/sec while on friction-free roller skates. What is the girl's velocity (speed and direction)?
    Answer

    pi = pf     p1i + p2i = p1f + p2f     m1v1i + m2v2i = m1v1f + m2v2f

    In this case: 0 = m1v1f + m2v2f

    v2f = -m1v1f/m2

    v2f = -[(1.5 kg)(1 m/sec)]/(50 kg) = -0.03 m/sec

    v2f = 0.03 m/sec away from the book

    Numerical Answers
    -0.03 m/sec
  17. Conservation of Energy and Momentum Problems

  18. Two billiard balls are traveling toward each other. A 500g billiard ball going 0.15 m/sec collides head-on and elastically with a 350 g billiard ball going 0.2 m/sec. What are the final velocities of each ball?
    Answer

    Conservation of Energy KEi + PEi = KEf + PEf + W

    Set h = 0 at the height of the billiard balls. All PE = 0. W = 0 (elastic).

    ½m1v1i² + ½m2v2i² = ½m1v1f² + ½m2v2f²   m1v1i² + m2v2i² = m1v1f² + m2v2f²

    (0.5 kg)(0.15 m/sec)² + (0.35 kg)(-0.2 m/sec)² = 0.02525 J = m1v1f² + m2v2f²

    0.02525 = (0.5)v1f² + (0.35)v2f²   (From Cons. of Energy, two unknowns)

    Conservation of Momentum   m1v1i + m2v2i = m1v1f + m2v2f

    (0.5 kg)(0.15 m/sec) - (0.35 kg)(0.2 m/sec) = 0.005 kg-m/sec = m1v1f + m2v2f

    0.005 = (0.5)v1f + (0.35)v2f   (From Cons. of Momentum, two unknowns)

    Substitute v2f = (0.005 - (0.5)v1f)/(0.35) into Cons. of Energy Eqn.

    0.02525 = (0.5)v1f² + (0.35)[(0.005 - (0.5)v1f)/(0.35)]²

    (1.2143)v1f² - (0.01429)v1f - 0.02518 = 0   v1f = -0.15 m/sec or 0.138 m/sec

    v1f = -0.15 m/sec and v2f = 0.229 m/sec

    Numerical Answers
    -0.15 m/sec 0.229 m/sec
  19. A 10 g bullet is shot horizontally at a 1.5 kg wooden block suspended from the ceiling. The bullet is embedded into the wooden block. The bullet and block then swing in an arc to a height of 12 cm. What is the velocity of the bullet before hitting the wooden block? How much energy was given off when the bullet was embedded into the wooden block?
    Answer

    See webpage for more details. Set h = 0 at bottom.

    After bullet is in the block to maximum height.

    ½(m1 + m2)vi² + 0 = 0 + (m1 + m2)ghf + 0

    vi = [(2)([m1 + m2]ghf)/(m1 + m2)]½ = [(2)(9.8 m/sec²)(0.12 m)]½

    = 1.534 m/sec   The velocity just after the bullet is embedded in the block.

    Before the bullet hits to just after the bullet is embedded in the block.

    Use conservation of momentum:

    (0.01 kg)v1i + 0 = (1.501 kg)(1.534 m/sec)

    v1i = (1.501 kg)(1.534 m/sec)/(0.01 kg) = 230.2 m/sec

    Before bullet is in the block to maximum height.

    To get W use conservation of energy from before collision to maximum height.

    ½(0.01 kg)(230.2 m/sec)² + 0 + 0 + 0 = 0 + (1.51 kg)(9.8 m/sec²)(0.12 m) + W

    W = 529.9 J - 1.8 J = 528.1 J

    Numerical Answers
    230.2 m/sec 528.1 J
  20. A 300 g billiard ball going 0.5 m/sec collides head-on and elastically with a stationary ball that has three times the mass. What are the final velocities of each ball?
    Answer

    From conservation of energy when W = 0 (elastic collision), initial and final height is at zero, m2 = 3m1, v2i = 0:

    ½m1v1i² + 0 + 0 + 0 = ½m1v1f² + ½(3m1)v2f² + 0 + 0 + 0

    v1i² = v1f² + 3v2f²     (0.5)² = v1f² + 3v2f²

    From conservation of momentum:

    m1v1i + 0 = m1v1f + 3m1v2f

    v1i = v1f + 3v2f     0.5 = v1f + 3v2f   v1f = 0.5 - 3v2f

    Substitute v1f into the energy equation, and then solve for v2f:

    (0.5)² = (0.5 - 3v2f)² + 3v2f² = (0.5)² - (2)(0.5)(3v2f) + (3v2f)² + 3v2f²

    0 = -3v2f + 12v2f² = (3)(-1 + 4v2f)(v2f)

    The two roots of the equation are v2f = 0 (physically impossible), or v2f = 0.25.

    v2f = 0.25 m/sec and v1f = 0.25 - 3v2f = 0.25 - 3(0.25) = -0.5 m/sec

    Numerical Answers
    0.25 m/sec -0.5 m/sec
  21. Conservation of Energy with Rotational Kinetic Energy

  22. A 20 g disk with a radius of 1 cm is released from rest at the top of a smooth inclined plane, where the height from the bottom of the incline is 12 m. The disk reaches the bottom of the incline with a speed of 12 m/sec. What is the numerical value of the moment of inertia for this disk?
    Answer

    Conservation of Energy KEit + KEir + PEi = KEft + KEfr + PEf + W

    ½mvi² + ½Iωi² + mghi = ½mvf² + ½Iωf² + mghf  

    Set h = 0 at the bottom of the incline. W = 0, vi = 0, ωi = 0, and hf = 0.

    0 + 0 + mghi = ½mvf² + ½Iωf² + 0  

    I = [mghi - ½mvf²]/[½ωf²]   ; ωf = vf/r = (12 m/sec)/(0.01 m) = 1200 rad/sec

    I = [(0.02 kg)(9.8 m/sec²)(12 m) - ½(0.02 kg)(12 m/sec)²]/[½(1200 rad/sec)²]

    I = 1.27 x 10-6 kg-m²

    Numerical Answers
    1.27 x 10-6 kg-m²
  23. What will the final speed be when a hollow sphere (I = ⅔mr²) is released from rest at a height of 80 cm and rolls down a 30 degree smooth incline?
    Answer

    Conservation of Energy KEit + KEir + PEi = KEft + KEfr + PEf + W

    0 + 0 + mghi = ½mvf² + ½Iωf² + 0 + 0

    ghi = ½vf² + ½[⅔r²][vf/r]² = ½vf² + ⅓vf² = ⅚vf²

    vf = [(6/5)ghi]½

    vf = 3.07 m/sec

    Numerical Answers
    3.07 m/sec
  24. EXTRA PROBLEMS
  25. A 300 g billiard ball going 0.10 m/sec collides head-on and elastically with a 350 g billiard ball going 0.13 m/sec in the opposite direction. What are the final velocities of each ball?
  26. What are the final momentums and velocities after a 0.22 kg air hockey puck initially traveling 0.4 m/sec at 45° collides with a 0.03 kg puck traveling at 0.3 m/sec at -30° (330° from the x-axis)?
  27. An airtrack car collides head-on and elastically with a second stationary car that has the same mass as the first car. Mathematically show that the final velocity of the first car is zero and that the final velocity of the second car is the same as the initial velocity of the first car.
    Answer

    From conservation of energy: W = 0 (elastic collision), initial and final height is along the track where we assign h equal to zero and v2i = 0:

    ½m1v1i² + 0 + 0 + 0 = ½m1v1f² + ½(m2)v2f² + 0 + 0 + 0

    the one-half and the mass will cancel out, so

    v1i² = v1f² + v2f²

    From conservation of momentum:

    m1v1i + 0 = m1v1f + m2v2f   but m1 = m2, so

    v1i = v1f + v2f     v1f = v1i - v2f   and substituting into energy equation:

      v1i² = (v1i - v2f)² + v2f²   =   v1i² - 2v1iv2f + v2f² + v2f²

      0 = 2v2f(v2f - v1i)

    The two roots of the equation are v2f = 0 (physically impossible), or v2f = v1i.

      v2f = v1i

    Answer

    From conservation of energy: W = 0 (elastic collision), initial and final height is along the track where we assign h equal to zero and v2i = 0:

    ½m1v1i² + 0 + 0 + 0 = ½m1v1f² + ½(m2)v2f² + 0 + 0 + 0

    the one-half and the mass will cancel out, so

    v1i² = v1f² + v2f²

    From conservation of momentum:

    m1v1i + 0 = m1v1f + m2v2f   but m1 = m2, so

    v1i = v1f + v2f     v2f = v1i - v1f   and substituting into energy equation:

      v1i² = v1f² + (v1i - v1f)²   =   v1f² + v1i² - 2v1iv1f + v1f²

      0 = 2v1f(v1f - v1i)

    The two roots of the equation are v1f = v1i (physically impossible), or v1f = 0.

      v1f = 0

Torque Problems

  1. A 2 m long, 20 N uniform board is leaned against a smooth vertical wall, creating a 40° angle between the wall and the board. If the frictional force at the bottom of the board is 15 N, how high up the board, from the floor, could a 10 N box be placed without the board slipping?
    Answer

    L = 2 m
    Fb = 20 N
    Fbx = (20 N)sin50° = 15.32 N
    Fby = (20 N)cos50° = 12.86 N

    Fmx = (10 N)sin50° = 7.66 N
    Fmy = (10 N)cos50° = 6.43 N
    Fwy = (15 N)sin50° = 11.49 N

    Forces in x: Fw = Ff = 15 N
    Forces in y: Fv = 20 N + 10 N = 30 N

    Set l = 0 at the bottom.   τin = τout     (Fwy)lw = (Fby)lb + (Fmy)lm

      (11.49 N)(2 m) = (12.86 N)(1 m) + (6.43 N)(x)   x = 1.57 m

    Numerical Answers
    1.57 m
  2. A straight, uniform ladder that is 2.5 m long is positioned against a wall. If both the wall has an upward frictional force of 15 N and the ground provides 20 N of horizontal frictional force and the ladder weighs 40 N, how high on the ladder can a 600 N person climb if the ladder makes a 35° angle with the wall?
    Answer

    L = 2.5 m; θ=35°
    Fb = 40 N
    Fbx = (40 N)sin55°
    = 32.77 N
    Fby = (40 N)cos55°
    = 22.94 N

    Fm = 600 N
    Fmx = (600 N)sin55° = 491.49 N
    Fmy = (600 N)cos55° = 344.15 N
    Fwy = (20 N)sin55° = 16.38 N
    Fwf = 15 N (An upward frictional force at the wall, not shown in diagram.)
    Fwfy = (15 N)cos35° = 12.28 N

    Forces in x: Fw = Ff = 20 N
    Forces in y: Fv + Fwfy = 40 N + 600 N = 640 N

    Set l = 0 at the bottom.   τin = τout     (Fwy)lw + (Fwfy)lwf = (Fby)lb + (Fmy)lm

      (16.38 N)(2.5 m) + (12.28 N)(2.5 m) = (22.94 N)(1.25 m) + (344.15 N)(x)
      x = 0.125 m

    Numerical Answers
    0.125 m
  3. A 3m, uniform 50 N plank hangs 0.6 m over the side of a pirate ship. What is the maximum weight that a person can have without toppling the plank when standing on the end?
    Answer

    Fb = 50 N;   Fm = ?
    Forces in y: Fv = Fb + Fm

    Set l = 0 at the Fv.

    τin = τout     (Fm)lm = (Fb)lb   (Fm)(0.6 m) = (50 N)(0.9 m)

      Fm = 75 N

    Numerical Answers
    75 N
  4. Where is the center of mass of a nonuniform, 10 kg, 3 m long board that hangs 0.80 m over the edge of a platform while holding a maximum weight of 500 N on the extended end of the board?
    Answer

    Fb = (10 kg)(9.8 m/sec²) = 980 N;   Fm = 500 N
    Forces in y: Fv = Fb + Fm

    Set l = 0 at the Fv.

    τin = τout     (Fm)lm = (Fb)lb   (500 N)(0.8 m) = (980 N)(x)  
      x = 0.408 m

      xcm = 1.208 m from the end with the hanging mass.

    Numerical Answers
    1.208 m from the end with the hanging mass.
  5. What are the tensions, T1 and T2, in the supporting wires shown in Figure 7.2?
    Answer

    5 kg   (5 kg)(9.8 m/sec²) = 49 N
    8 kg   (8 kg)(9.8 m/sec²) = 78.4 N
    Forces in y: T1 + T2 = 49 N + 5 N + 78.4 N = 132.4 N

    Set l = 0 at the T2.

    τin = τout     (T1)(0.3 m) + (78.4 N)(0.15 m) = (5 N)(0.15 m) + (49 N)(0.45 m)   T1 = 36.8 N

      T2 = 132.4 N - 36.8 N = 95.6 N

    Numerical Answers
    T1 = 36.8 N   T2 = 95.6 N
  6. What is the tension in the horizontal wire connected to the wall in Figure 7.3 for a uniform supporting rod?
    Answer

    25 kg   (25 kg)(9.8 m/sec²) = 245 N
    Set the x-axis along the supporting rod. Tx = Tsin50° and Ty = Tcos50°

    Set l = 0 at the bottom of the supporting rod.

    τin = τout     [(20 N)cos40°](L/2) + [(245 N)cos40°](L) = (Ty)(L)
      Ty = 195.3 N

      T = Ty/cos50° = 303.9 N

    Numerical Answers
    303.9 N
  7. What is the tension shown in Figure 7.3 if the center of mass of the supporting rod is one-fourth of the way up from the wall, instead of at the center of the rod?
    Answer

    25 kg   (25 kg)(9.8 m/sec²) = 245 N
    Set the x-axis along the supporting rod. Tx = Tsin50° and Ty = Tcos50°

    Set l = 0 at the bottom of the supporting rod.

    τin = τout     [(20 N)cos40°](L/4) + [(245 N)cos40°](L) = (Ty)(L)
      Ty = 191.5 N

      T = Ty/cos50° = 297.9 N

    Numerical Answers
    297.9 N
  8. EXTRA PROBLEMS
  9. Two children are sitting on either end of a seesaw. If one person weighs three-fourths as much as the other, where must they each sit in order to balance?
  10. Two students are moving a 500 N instrument across the room by lifting opposite ends of the instrument. If the center of mass of the instrument is one-fourth of the distance from one end, how much force will each student exert?
  11. A 30 kg uniform board that is 2 m long is placed on a horizontal table. Part of the length of the board overhangs the table. A 50 kg lamp is placed on the end of the board that extends out from the table. What is the maximum distance that the board can overhang and still stay on the table?
  12. A child's mobile is made by hanging a 2 g star and a 3 g moon on opposite ends of a 20 g uniform rod and hanging the rod with some string. If the rod is 30 cm long and hangs horizontally, where is the string attached?
  13. A 5 m rod has a 30 N mass hanging from one end. A fulcrum is placed 3 m from the end that has the 30 N mass. The center of mass of the rod is also 3 m from the 30 N mass. A second mass is placed at the opposite end of the rod from the 30 N mass. (a) What is the weight of the second mass? (b) If the rod is 50 N, what force does the fulcrum exert?
  14. A uniform drawbridge has a weight of 6000 N. The cables are attached three-fourths of the way along the bridge from the castle and are at a 45° angle when the drawbridge is horizontal. (a) How much tension must the cables be able to handle when the bridge is horizontal? (b) How much tension must the cables be able to handle when the bridge is horizontal and a 600 N person is standingat the end of the drawbridge opposite the castle? (c) Determine the maximum tension on the cables as the drawbridge is raised. (d) Show how the vertical force exerted by the castle on the bridge where they connect changes as the drawbridge is raised.