Physics I Homework Page

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Homework Without Visible Answers

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This page has all of the homework with answers 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?

Horizontal Acceleration (Friction)

4. 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   v_o = 8.94 m/sec   s = ?   a = -6.86 m/sec²   t = ?

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

   v² = v_o² + 2as;  

   s = [v² - v_o²]/2a = [0 - (8.94 m/sec)²]/(2)(-6.86 m/sec²) = 5.8 m

5. 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   v_o = ? m/sec   s = 30 m   a = -3.43 m/sec²   t = ?

   v² = v_o² + 2as;

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

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

6. 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   v_o = ? m/sec   s = ? m   a = -4.86 m/sec²   t = 2.1 sec

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

   v² = v_o² + 2as;  

   s = [v² - v_o²]/2a = [0 - (10.2 m/sec)²]/(2)(-4.86 m/sec²) = 10.7 m

Vertical Acceleration (Gravity)

7. 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   v_o = ? m/sec   s = ? m   a = -9.8 m/sec²   t = 2.0 sec

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

   v² = v_o² + 2as;

   s = [v² - v_o²]/2a = [(-20 m/sec)² - (-0.4 m/sec)²]/(2)(-9.8 m/sec²) = -20.4 m

8. 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   v_o = ? m/sec   s = 10 m   a = -9.8 m/sec²   t = ? sec

   v² = v_o² + 2as; v_o = [v² - 2as]^½ = [0 - (2)(-9.8 m/sec²)(10 m)]^½ = 14 m/sec

   v = v_o + at;   t = (v - v_o)/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

9. 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   v_o = 0 m/sec   s = ? m   a = -9.8 m/sec²   t = 4.0 sec

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

   The well is 78.4 m deep.

   Also:

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

10. 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   v_o = 0 m/sec   s = ? m   a = -9.8 m/sec²   t = 3.0 sec

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

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

    s = v_ot + 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!

11. 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   v_o = 0 m/sec   s = ? m   a = -9.8 m/sec²   t = ? sec

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

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

    It dropped 127.5 m.

Horizontal and Vertical Acceleration (Projectiles)

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

12. A frog jumps off of a 0.5 m bank at 25^o 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: v_oy = v_osin25° = (3 m/sec)(sin25°) = 1.27 m/sec

    In x: v_ox = v_ocos25° = (3 m/sec)(cos25°) = 2.72 m/sec

    	y	x
    v	? m/sec	2.72 m/sec
    vo	1.27 m/sec	2.72 m/sec
    s	-0.5 m	? m
    a	-9.8 m/sec	0 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.

    v_y = [v_oy² + 2a_ys_y]^½ = [(1.27 m/sec)² + (2)(-9.8 m/sec²)(-0.5 m)]^½

    = -3.38 m/sec

    t = (v_y - v_oy)/a_y = [(-3.38 m/sec) - 1.27 m/sec]/(-9.8 m/sec²) = 0.474 sec

    s_x = v_oxt + ½at² = (2.72 m/sec)(0.474 sec) + 0 = 1.29 m

13. 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

    	y	x
    v	? m/sec	? m/sec
    vo	0 m/sec	? m/sec
    s	-1000 m	2000 m
    a	-9.8 m/sec	0 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.

    v_y = [v_oy² + 2a_ys_y]^½ = [0 + (2)(-9.8 m/sec²)(-1000 m)]^½

    = -140 m/sec

    t = (v_y - v_oy)/a_y = [(-140 m/sec) - 0]/(-9.8 m/sec²) = 14.3 sec

    v_ox = s_x/t = (2000 m)/(14.3 sec) = 140 m/sec

    For the final velocity:

    v = [(v_x)² + (v_y)²]^½ = [(-140 m/sec)² + (140 m/sec)²]^½ = 198 m/sec

    θ = tan^-1[(v_x)/v_y)] = tan^-1(140/140) = 45°

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

14. 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 = [(v_ox)² + (v_oy)²]^½ = [(160 m/sec)² + (75 m/sec)²]^½ = 176.7 m/sec

    θ = tan^-1[(v_oy)/v_ox)] = tan^-1(75/160) = 25.1°

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

    	y	x
    v	0 m/sec	160 m/sec
    vo	75 m/sec	160 m/sec
    s	? m	? m
    a	-9.8 m/sec	0 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.

    s_y = [(v_y)² - (v_oy)²)]/[2a_y] = [(0 - (75 m/sec)²)]/[2(-9.8 m/sec²)] = 287 m

    t = (v_y - v_oy)/a_y = [(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): s_x = v_ot = (160 m/sec)(15.3 sec) = 2449 m

15. When a spring gun is held at an angle of 40^o 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 (a)

    	y	x
    v	vsinθ m/sec	vcosθ m/sec
    vo	vosinθ m/sec	vocosθ m/sec
    s	0 m	30 m
    a	-9.8 m/sec	0 m/sec
    t	t sec	t 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 = (v_osinθ m/sec)(t) or t = (30 m)/(v_osinθ m/sec)
    In y: s_y = v_oyt + ½a_yt² = [v_oy + ½a_yt]t

    0 = [v_oy + ½a_yt]t says that either t = 0 or [v_oy + ½a_yt] = 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 = -v_oy/½a = -(v_osinθ)/½a

    That means:

    -(v_osinθ)/½a = (30 m)/(v_osinθ m/sec) or -(v_o²)(sinθ)(cosθ) = ½a(30)

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

    = 17.3 m/sec

    Answer (b and c)

    From part (a):

    v_ox = v_ocosθ m/sec = (17.3 m/sec)cos40° = 13.3 m/sec

    v_oy = v_osinθ = (17.3 m/sec)sin40° = 11.12 m/sec

    Make a table going from start to max.

    	y	x
    v	0 m/sec	vx = vox m/sec
    vo	11.12 m/sec	13.3 m/sec
    s	? m	15 m
    a	-9.8 m/sec	0 m/sec
    t	t sec	t sec

    s_y = [(v_y)² - (v_oy)²)]/[2a_y] = [(0 - (11.2 m/sec)²)]/[2(-9.8 m/sec²)]

    = 6.4 m

    t = (v_y - v_oy)/a_y = [(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

16. An object is thrown from a bridge at an angle of 40^o 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.

    	y	x
    v	vsinθ m/sec	vx = vox m/sec
    vo	vosinθ m/sec	vocosθ m/sec
    s	-15 m	10 m
    a	-9.8 m/sec	0 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)/(v_ocos40°)

    s_y = (v_oy)t + ½at²

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

    -15 m = (v_o)(sin40°)[(10 m)/(v_ocos40°)] + ½(-9.8 m/sec²)[(10 m)/(v_ocos40°)]²

    v_o = 5.975 m/sec; v_oy = 3.84 m/sec; v_ox = 4.58 m/sec;

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

    v_y = 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°

    t = 2.18 sec, v = -17.52 m/sec at 75.3° below the positive x-axis.

17. An archer shoots an arrow at an angle of 10^o 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

    	y	x
    v	0 m/sec	vx = vox m/sec
    vo	? m/sec	? m/sec
    s	? m	? m
    a	-9.8 m/sec	0 m/sec
    t	1.05 sec	1.05 sec

    Symmetry makes this problem much easier.

    Make a table going from start to max.

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

    v_o = (v_oy)/sinθ = (10.29 m/sec)/sin10° = 59.3 m/sec at 10°

    v_ox = (v_o)(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 = [(v_x)² + (v_y)²]^½ = 59.3 m/sec at -10°

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

EXTRA PROBLEMS

18. 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]
19. 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?
20. 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.
21. 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?
22. 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?
23. 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

    

24. 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. [v_oy = 34 m/sec, v_o = 68 m/sec, s_x = 588.89 m]
25. 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?
26. I was recently reading an article about calculations for how far a baseball travels when hit by a bat. In that article they made the following statement:

    "Your students might immediately want to plug the launch speed and launch angle into the range equation,

    R=(v₀² sin2θ)/g,

    where v₀ is the launch speed, g is the acceleration due to gravity, and θ is the vertical launch angle."

    R is the range, or the distance in x. My questions is: can you derive this equation using our method of doing these problems?

    Answer

    

Net Force Problems

10. 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   = F_g2 - F_g1 = 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   = F_g2 - T₂ = ma

    T₂   = F_g2 - ma = (0.4 kg)(9.8 m/sec²) - (0.4 kg)(4.45 m/sec²)

    T₂   = 2.14 N

11. 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   = F_g2 = (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   = F_g2 - T = m₂a

    T   = F_g2 - m₂a = (2 kg)(9.8 m/sec²) - (2 kg)(3.92 m/sec²)

    T   = 11.76 N

12. 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

    F_gx = F_gsinθ ;   F_gy = F_gcosθ ;   F_n = F_gy ;   f_k = μ_kF_n

    m = (20 N)/(9.8 m/sec²) = 2.04 kg

    x-axis (along the incline):

    netForce   = F_gx - f_k = ma

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

    y-axis (perpendicular to the plane):

    f_k = μ_kF_n

    μ_k = f_k/F_n = (5.22 N)/[(20 N)cos30°]

    μ_k = 0.3

13. 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

    F_5x = F₅sinθ = (49 N)sin25° = 20.71 N ; F_5y = F₅cosθ = (49 N)cos25° = 44.41 N

    F_n = F_5y = 44.41 N ;   f_k = μ_kF_n = (0.11)(44.41 N) = 4.88 N

    Whole System:

    netForce   = F₁₂ - f_k - F_5x = (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

Centripetal Force Problems

14. 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

    f_s = F_c ;   f_s = μ_sF_n ;   F_c = mv²/r ; F_n = mg

    ⇒   μ_smg = mv²/r   ⇒ μ_sg = v²/r

    ⇒   v = [rμ_sg]^½ = [(0.5)(9.8 m/sec²)(20 m)]^½ = 9.9 m/sec

15. 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.

    F_gx = F_cx

    F_gsinθ = F_ccosθ

    ⇒   tanθ = F_c/F_g = (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°

16. 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: F_g = F_c.

    ⇒   mg = mv²/r

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

EXTRA PROBLEMS
17. 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?
18. 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?
19. Calculate the acceleration when a 20 N sled is being pulled over smooth ice with a horizontal force of 3 N.
20. 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.
21. 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?
22. 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

    

23. If F = 40 N, T₁ = 20 N, and T₂ = 30 N, what are the angles θ₁ and θ₂?

    Answer

    

24. 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

    

25. 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 μ_s = 0.7?

    Answer

    

26. 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 μ_s = 0.7? What if μ_s 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

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 = F_xs_x = [(1.5 kg)(9.8 m/sec²)(sin20°)](2 m) = 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

Conservation of Energy Problems

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

   KE_i + PE_i = KE_f + PE_f + W   ⇒   ½mv_i² + mgh_i = ½mv_f² + mgh_f + Fs

   Set h = 0 at the bottom of the incline.

   ⇒ ½m(7 m/sec)² + m(9.8 m/sec²)[(2 m)(sin20°)] = ½mv_f² + 0 + 0

   The mass cancels out

   ⇒ v_f² = 2{½(7 m/sec)² + (9.8 m/sec²)[(2 m)(sin20°)]}

   ⇒ v_f = (62.41 m²/sec²)^½ = 7.9 m/sec

5. 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

   KE_i + PE_i = KE_f + PE_f + W   ⇒   ½mv_i² + mgh_i = ½mv_f² + mgh_f + Fs

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

   ⇒ 0 + m(9.8 m/sec²)(4 m) = ½mv_f² + 0 + 0

   The mass cancels out

   ⇒ v_f = [2{9.8 m/sec²)(4 m)]^½ = 8.85 m/sec

6. What is the final velocity when a 3.3 kg mass slides down a 50° incline if: v_i = 0, s = 3.3 m, µ_k = 0.12?
   Answer

   KE_i + PE_i = KE_f + PE_f + W   ⇒   ½mv_i² + mgh_i = ½mv_f² + mgh_f + W

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

   ⇒ W = f_k(s) = µ_kF_n(s) = µ_kF_gy(s) = µ_k(mgcosθ)(s)

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

   ⇒ 0 + (9.8 m/sec²)(2.528 m) = ½v_f² + 0 + (0.12)(9.8 m/sec²)(cos50°)(3.3 m)

   ⇒ 0 + 24.774 m²/sec² = ½v_f² + 0 + 2.495 m²/sec²

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

   

7. 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?
   Answers:

   Part (a):

   KE_i + PE_i = KE_f + PE_f + W   ⇒   ½mv_i² + mgh_i = ½mv_f² + mgh_f + Fs

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

   F_g = mg   F_gx = mgsinθ

   With W = 0 the mass will cancel out.

   ⇒ 0 + (9.8 m/sec²)(1.5 m) = ½v_f² + 0 + 0

   ⇒ v_f = [2(9.8 m/sec²)(1.5 m)]^½ = 5.42 m/sec

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

   KE_i + PE_i = KE_f + PE_f + W   ⇒   ½mv_i² + mgh_i = ½mv_f² + mgh_f + W

   Set h = 0 at the bottom of the incline. h_i = h_f = 0.   v_i = 5.42 m/sec.   v_f = 0.

   W = Fs = f_ks = F_nµ_ks = F_gµ_ks = (mg)(µ_k)(s)

   ⇒ ½mv_i² + 0 = 0 + 0 + (mg)(µ_k)(s)

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

   ⇒ v_i² + 0 = 0 + 0 + (2)(g)(µ_k)(s)

   ⇒ s = (v_i²)/[(2)(g)(µ_k)] = (5.42 m/sec)²/[(2)(9.8 m/sec²)(0.11)] = 13.6 m

   

8. 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

   KE_i + PE_i = KE_f + PE_f + W   ⇒   ½mv_i² + mgh_i = ½mv_f² + mgh_f + Fs

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

   v_ix = (20 m/sec)(cos30°) = 17.32 m/sec   v_iy = (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) = ½v_fy² + 0 + 0

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

   v_f = [v_fx² + v_fy²]^½ = [(17.32 m/sec)² + (19.8 m/sec)²]^½ = 26.3 m/sec

9. 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

   KE_i + PE_i = KE_f + PE_f + W   ⇒   ½mv_i² + mgh_i = ½mv_f² + mgh_f + Fs

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

   v_i = (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 = ½v_f² + (9.8 m/sec²)(5 m) + 0

   ⇒ v_f = [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:   v_f = 0   ⇒   v_i = [2gh_f]^½ = 9.9 m/sec

10. 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

    KE_i + PE_i = KE_f + PE_f + W   ⇒   ½mv_i² + mgh_i = ½mv_f² + mgh_f + Fs

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

    v_i = 0.   v_f = 1 m/sec.   m = F_g/g = (500 N)/(9.8 m/sec²) = 51 kg

    W = Fs = (F_gx - f_k)s = (F_gsinθ - F_nµ_k)s = (F_gsinθ - F_gyµ_k)s

    = (F_gsinθ - [F_gcosθ]µ_k)s = F_g[sinθ]s - F_g[cosθ]µ_ks

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

    ⇒ 0 + mgh_i = ½mv_f² + 0 + W

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

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

Conservation of Momentum Problems

11. 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

    p_i = p_f   ⇒   p_1i + p_2i = p_1f + p_2f   ⇒   m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f

    In this case: m₁v_1i + m₂v_2i = (m₁ + m₂)v_f

    ⇒ v_f = (m₁v_1i + m₂v_2i)/(m₁ + m₂) = [(0.4 kg)(0.12 m/sec) + 0]/(0.4 kg + 0.35 kg)

    ⇒ v_f = 0.064 m/sec

12. 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

    p_i = p_f   ⇒   p_1i + p_2i = p_1f + p_2f   ⇒   m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f

    In this case: m₁v_1i + m₂v_2i = (m₁ + m₂)v_f

    ⇒ v_f = (m₁v_1i + m₂v_2i)/(m₁ + m₂)

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

    ⇒ v_f = 0.036 m/sec

13. 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

    p_i = p_f   ⇒   p_1i + p_2i = p_1f + p_2f   ⇒   m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f

    In this case: 0 = m₁v_1f + m₂v_2f

    ⇒ v_2f = -m₁v_1f/m₂

    ⇒ v_2f = -[(55 kg)(2 m/sec)]/(45 kg) = -2.44 m/sec

    ⇒ v_2f = 2.44 m/sec away from the boy

14. 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

    p_i = p_f   ⇒   p_1i + p_2i = p_1f + p_2f   ⇒   m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f

    In this case: 0 = m₁v_1f + m₂v_2f

    ⇒ v_2f = -m₁v_1f/m₂

    ⇒ v_2f = -[(1.5 kg)(1 m/sec)]/(50 kg) = -0.03 m/sec

    ⇒ v_2f = 0.03 m/sec away from the book

15. Two giants are playing modified ice hockey (no friction). One giant hits a 45 kg puck so it is moving at 2 m/sec along a positive x-axis. A second giant, who is over and down a ways, smacks a 60 kg puck directly up the y-axis at 3 m/sec. The two pucks collide, stick together and remain together as they move. Determine the final velocity (magnitude and direction) of the combined pucks.
    Answer

Conservation of Energy and Momentum Problems

16. 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 ⇒ KE_i + PE_i = KE_f + PE_f + W

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

    ⇒ ½m₁v_1i² + ½m₂v_2i² = ½m₁v_1f² + ½m₂v_2f²   ⇒ m₁v_1i² + m₂v_2i² = m₁v_1f² + m₂v_2f²

    ⇒ (0.5 kg)(0.15 m/sec)² + (0.35 kg)(-0.2 m/sec)² = 0.02525 J = m₁v_1f² + m₂v_2f²

    ⇒ 0.02525 = (0.5)v_1f² + (0.35)v_2f²   (From Cons. of Energy, two unknowns)

    Conservation of Momentum ⇒   m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f

    ⇒ (0.5 kg)(0.15 m/sec) - (0.35 kg)(0.2 m/sec) = 0.005 kg-m/sec = m₁v_1f + m₂v_2f

    ⇒ 0.005 = (0.5)v_1f + (0.35)v_2f   (From Cons. of Momentum, two unknowns)

    Substitute v_2f = (0.005 - (0.5)v_1f)/(0.35) into Cons. of Energy Eqn.

    ⇒ 0.02525 = (0.5)v_1f² + (0.35)[(0.005 - (0.5)v_1f)/(0.35)]²

    ⇒ (1.2143)v_1f² - (0.01429)v_1f - 0.02518 = 0   ⇒ v_1f = +0.15 m/sec or -0.138 m/sec

    ⇒ v_1f = -0.138 m/sec and v_2f = 0.211 m/sec

17. 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)v_i² + 0 = 0 + (m₁ + m₂)gh_f + 0

    ⇒ v_i = [(2)([m₁ + m₂]gh_f)/(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)v_1i + 0 = (1.501 kg)(1.534 m/sec)

    ⇒ v_1i = (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 = 263.2 J

18. 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, v_2i = 0:

    ½m₁v_1i² + 0 + 0 + 0 = ½m₁v_1f² + ½(3m₁)v_2f² + 0 + 0 + 0

    v_1i² = v_1f² + 3v_2f²   ⇒   (0.5)² = v_1f² + 3v_2f²

    From conservation of momentum:

    m₁v_1i + 0 = m₁v_1f + 3m₁v_2f

    v_1i = v_1f + 3v_2f   ⇒   0.5 = v_1f + 3v_2f   ⇒ v_1f = 0.5 - 3v_2f

    Substitute v_1f into the energy equation, and then solve for v_2f:

    ⇒ (0.5)² = (0.5 - 3v_2f)² + 3v_2f² = (0.5)² - (2)(0.5)(3v_2f) + (3v_2f)² + 3v_2f²

    ⇒ 0 = -3v_2f + 12v_2f² = (3)(-1 + 4v_2f)(v_2f)

    The two roots of the equation are v_2f = 0 (physically impossible), or v_2f = 0.25.

    v_2f = 0.25 m/sec and v_1f = 0.25 - 3v_2f = 0.25 - 3(0.25) = -0.5 m/sec

Conservation of Energy with Rotational Kinetic Energy

19. 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 ⇒ KE_it + KE_ir + PE_i = KE_ft + KE_fr + PE_f + W

    ⇒ ½mv_i² + ½Iω_i² + mgh_i = ½mv_f² + ½Iω_f² + mgh_f  

    Set h = 0 at the bottom of the incline. W = 0, v_i = 0, ω_i = 0, and h_f = 0.

    ⇒ 0 + 0 + mgh_i = ½mv_f² + ½Iω_f² + 0  

    ⇒ I = [mgh_i - ½mv_f²]/[½ω_f²]   ; ω_f = v_f/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

20. 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 ⇒ KE_it + KE_ir + PE_i = KE_ft + KE_fr + PE_f + W

    ⇒ 0 + 0 + mgh_i = ½mv_f² + ½Iω_f² + 0 + 0

    ⇒ gh_i = ½v_f² + ½[⅔r²][v_f/r]² = ½v_f² + ⅓v_f² = ⅚v_f²

    ⇒ v_f = [(6/5)gh_i]^½

    ⇒ v_f = 3.07 m/sec

    Numerical Answers

    3.07 m/sec

EXTRA PROBLEMS
21. 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?
22. 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)?
23. 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 v_2i = 0:

    ½m₁v_1i² + 0 + 0 + 0 = ½m₁v_1f² + ½(m₂)v_2f² + 0 + 0 + 0

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

    v_1i² = v_1f² + v_2f²

    From conservation of momentum:

    m₁v_1i + 0 = m₁v_1f + m₂v_2f   but m₁ = m₂, so

    v_1i = v_1f + v_2f   ⇒   v_1f = v_1i - v_2f   and substituting into energy equation:

      ⇒ v_1i² = (v_1i - v_2f)² + v_2f²   =   v_1i² - 2v_1iv_2f + v_2f² + v_2f²

      ⇒ 0 = 2v_2f(v_2f - v_1i)

    The two roots of the equation are v_2f = 0 (physically impossible), or v_2f = v_1i.

      ⇒ v_2f = v_1i

    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 v_2i = 0:

    ½m₁v_1i² + 0 + 0 + 0 = ½m₁v_1f² + ½(m₂)v_2f² + 0 + 0 + 0

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

    v_1i² = v_1f² + v_2f²

    From conservation of momentum:

    m₁v_1i + 0 = m₁v_1f + m₂v_2f   but m₁ = m₂, so

    v_1i = v_1f + v_2f   ⇒   v_2f = v_1i - v_1f   and substituting into energy equation:

      ⇒ v_1i² = v_1f² + (v_1i - v_1f)²   =   v_1f² + v_1i² - 2v_1iv_1f + v_1f²

      ⇒ 0 = 2v_1f(v_1f - v_1i)

    The two roots of the equation are v_1f = v_1i (physically impossible), or v_1f = 0.

      ⇒ v_1f = 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
   F_b = 20 N
   F_bx = (20 N)sin50° = 15.32 N
   F_by = (20 N)cos50° = 12.86 N

   F_mx = (10 N)sin50° = 7.66 N
   F_my = (10 N)cos50° = 6.43 N
   F_wy = (15 N)sin50° = 11.49 N

   Forces in x: F_w = F_f = 15 N
   Forces in y: F_v = 20 N + 10 N = 30 N
   

   Set l = 0 at the bottom.   τ_in = τ_out   ⇒   (F_wy)l_w = (F_by)l_b + (F_my)l_m

   ⇒   (11.49 N)(2 m) = (12.86 N)(1 m) + (6.43 N)(x) ⇒   x = 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°
   F_b = 40 N
   F_bx = (40 N)sin55°
   = 32.77 N
   F_by = (40 N)cos55°
   = 22.94 N

   F_m = 600 N
   F_mx = (600 N)sin55° = 491.49 N
   F_my = (600 N)cos55° = 344.15 N
   F_wy = (20 N)sin55° = 16.38 N
   F_wf = 15 N (An upward frictional force at the wall, not shown in diagram.)
   F_wfy = (15 N)cos35° = 12.28 N

   Forces in x: F_w = F_f = 20 N
   Forces in y: F_v + F_wfy = 40 N + 600 N = 640 N

   Set l = 0 at the bottom.   τ_in = τ_out   ⇒   (F_wy)l_w + (F_wfy)l_wf = (F_by)l_b + (F_my)l_m

   ⇒   (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

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

   F_b = 50 N;   F_m = ?
   Forces in y: F_v = F_b + F_m

   Set l = 0 at the F_v.

   τ_in = τ_out   ⇒   (F_m)l_m = (F_b)l_b ⇒   (F_m)(0.6 m) = (50 N)(0.9 m)

   ⇒   F_m = 75 N

4. Where is the center of mass of a nonuniform, 100 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

   F_b = (100 kg)(9.8 m/sec²) = 980 N;   F_m = 500 N
   Forces in y: F_v = F_b + F_m

   Set l = 0 at the F_v.

   τ_in = τ_out   ⇒   (F_m)l_m = (F_b)l_b ⇒   (500 N)(0.8 m) = (980 N)(x)  
   ⇒   x = 0.408 m

   ⇒   x_cm = 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.

   τ_in = τ_out   ⇒   (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

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. T_x = Tsin50° and T_y = Tcos50°
   

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

   τ_in = τ_out   ⇒   [(20 N)cos40°](L/2) + [(245 N)cos40°](L) = (T_y)(L)
   ⇒   T_y = 195.3 N

   ⇒   T = T_y/cos50° = 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. T_x = Tsin50° and T_y = Tcos50°
   

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

   τ_in = τ_out   ⇒   [(20 N)cos40°](L/4) + [(245 N)cos40°](L) = (T_y)(L)
   ⇒   T_y = 191.5 N

   ⇒   T = T_y/cos50° = 297.9 N

EXTRA PROBLEMS
8. 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?
9. 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?
10. 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?
11. 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?
12. 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?
13. 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.