Section Test:Work

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1. You push against a brick wall for 30 seconds with a force of 90N for 30 seconds. How much work is done through this vigorous activity?

0 J
90 J
2700 J
81000 J
→ To know how much work is done, one MUST know the distance that the object was moved/pushed. In this case, we are assuming that you are unable to move the wall (it is a wall after all) and so no work is done. As Work = F*d, and distance here is zero!

2. A box rests on a horizontal, frictionless surface. A girl pushes on the box with a force of 18 N to the right and a boy pushes on the box with a force of 12 N to the left. The box moves 4.0 m to the right. The work done by the girl is ____ and the work done by the boy is _____.

72 J , 48 J
48 J , 72 J
72 J , -48 J
-48 J , 72 J
→ To solve this problem we must use the work formula for each person involved
→ W = Fd{cos}{theta}
→ For the girl:
→ W = Fd{cos}{theta} = (18 N)(4m)({cos}0) = 72 J
→ For the boy:
→ W = Fd{cos}{theta} = (18 N)(4m)({cos}180) = -48 J
→ Note that the block went the opposite direction that the boy was pushing, thus we could either make the distance negative, or make the angle 180 degrees (opposite), either way one needs to recognize that the value of his work is negative since there was actually more work against him!

3. A person pushes a 10 kg cart a distance of 20 meters by exerting a 60 N horizontal force however friction resists with a force of 50 N. How much kinetic energy does the cart have at the end of the 20 meters if it started from rest?

20 J
40 J
100 J
200 J
→ To answer this problem, one must remember that energy is always conserved, so the final kinetic energy must come from the work that is put in by the person pushing. That being said, resistance will "eat" up some of this work, and so only the extra energy that overcomes the resistance will be left to become kinetic energy.
→ {K.E.}_final = {W}_{net} = {W}_{person pushing} - {W}_{resistance}
→ {K.E.}_final = {W}_{net} = F_{person pushing} d cos(0) - {F}_{resistance} d cos(0)
→ {K.E.}_final = {W}_{net} = (60)(20)cos(0) - (50)(20)cos(0)
→ {K.E.}_final = 200 J
→ As a side, the weight of the cart (10kg) plays no role in answering this question as we are given the force of friction. Otherwise one would have to use it along with a coefficient of friction to determine the force.

4. Which of the following expressions correctly defines the unit of work?

{N}/{m}
{kg * m^2}{s^2}
{N}/{m * s}
{W}/{s}
→ The unit of work is the Joule, which is equal to Force x distance, or N*m. Force then, is (kg x m) / s. Joules can also be expressed using watts, as W*s (not W/s).

5. A boy release a 2 kg toy, from rest, at the top of a slide inclined at 20.0° above the horizontal. The coefficient of kinetic friction is 0.200. How much work does friction do along the length of the slide?

7 J
15 J
35 J
43 J
→ This problem is solved easiest if you draw the force triangle that is affecting the block on the incline.
→ Image:testwork_inclineplane.png
→ The force of gravity is straight down (the hypotenuse on the triangle)
→ F_{gravity} = m*a = (2{kg})*(9.8{m/s^2}) = 19.6 N
→ The normal force, which goes along the adjacent edge of the triangle, perpendicular to the plane, is equal to the component of gravity in that direction.
→ F_{normal} = F_{gravity} * cos(20) = 18.42 N
→ Be sure to understand why in this triangle, we used cos to get the normal force!
→ Finally, friction is equal to the normal force times the coefficient of friction
→ F_{friction} = F_{normal} * mu = 18.42 N * 0.20 = 3.68 N
→ Now, Work = F * d, therefore we have,
→Work = F * d = 3.68 N * 4.0 m = 14.73 J

6. What will the toy's speed be after sliding 4.00 m along the slide?

3.6 m/s
7.7 m/s
14.2 m/s
44.0 m/s
→ To determine how fast the toy will be sliding at the end of the incline, we first need to determine how much potential energy the toy had at the top, and then determine the kinetic energy it has keeping in mind to subtract the energy lost to friction.
→ {K.E.}_{at bottom} = {P.E.}_{at top} - {Work}_{friction}
→ {K.E.}_{at bottom} = mgh - {Work}_{friction}
→ {K.E.}_{at bottom} = (2 kg)(9.8 m/s^2)(4 sin(20)) - (14.73 J)
→ {K.E.}_{at bottom} = 58.94 J
→ Now we can determine the velocity.
→ {K.E.}_{at bottom} = 58.94 J = (1/2) * m * v^2 = (1/2) * (2 kg) * (v^2) = v^2
→ V = 7.68 m/s
→ Note that this problem could also be solved using regular translational motion formulas once one calculate the net force along the slope (Fg * sin(20) - Ffriction)

7. A 15-g bullet is accelerated in a rifle barrel 75 cm long to a speed of 400m/s. What is the average force that acts on the bullet while it is being accelerated?

300 N
900 N
1200 N
1600 N
→ To solve this we must first recognize that we will need to use W = Fd and solve for F. To use this however, we will need to calculate how much work was done first using an alternative formula, namely, we know how much K.E. the bullet has at the end of the barrel and thus the change in K.E. must be the amount of work that was put into making the bullet move!
→ Work = delta K.E. = {K.E.}_{final} - {K.E.}_{initial} = (1/2)(m)(v^2) - 0
→ Work = (1/2)(0.015 kg)(400^2) = 1200 J
→ Now that we have the total energy that went into work, we can solve for the average force that was applied along the distance d.
→ F = W/d = (1200 J)/(0.72 m) = 1600 N

8. An 85 g arrow is fired from a bow whose string exerts an average force of 95 N on the arrow over a distance of 80 cm. What is the speed of the arrow as it leaves the bow?

32 m/s
42 m/s
68 m/s
88 m/s
→ To solve this problem, we first need to calculate the work done by the bow on he arrow, and then use this energy to calculate the velocity since all of it will be transfered to the form of K.E.
→ {Work} = F * d = (95 N) * (0.8 m) = 76 J
→ {Work} = K.E. = (1/2)(m)(v^2)
→ v^2 = 2 * {Work} / m = 2 * (76 J) / (0.085 kg) = 1788
→ v = 42.28 m/s

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