Ten sense-making tasks on work, energy, and power. Explain your reasoning for every answer. The explanation counts as much as the answer.
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Each block is released from rest at the top of a frictionless track and slides to the bottom. The tracks have different shapes and lengths. The mass of each block and the height of each track are shown.
(a) Rank the blocks by their kinetic energy at the bottom, from greatest to least.
Greatest 1 ____ 2 ____ 3 ____ 4 ____ 5 ____ Least
(b) Rank the blocks by their speed at the bottom, from greatest to least.
Greatest 1 ____ 2 ____ 3 ____ 4 ____ 5 ____ Least
If any are equal, show the tie. If a ranking cannot be determined, say so. Explain your reasoning for both rankings.
Back to topA 20 kg box must be raised from the floor onto a shelf 1.0 m high. It can be lifted straight up, or pushed up a frictionless ramp 4.0 m long. Either way, the box moves at a constant slow speed.
Ava: "The ramp takes less work. That's the whole point of a ramp, you don't have to push as hard."
Ben: "The work is the same both ways. The ramp needs less force, but you push it over a longer distance."
Cam: "The ramp takes more work because the box travels 4.0 m instead of 1.0 m."
Which student(s), if any, do you agree with? Explain what is right or wrong about each statement. Calculate the work done by the person in each case. How would your answer change if the ramp had friction?
Back to topA ball is launched at 60° above the horizontal. Ignore air resistance. The system is the ball and Earth, with $U_g = 0$ at the launch height.
A student draws these energy bar charts for the launch and for the highest point:
What, if anything, is wrong with these bar charts? If something is wrong, explain the error and draw the correct charts. If nothing is wrong, explain why they are correct.
Back to topFor each situation, choose the sign of the work done by the named force on the object:
A. positive B. negative C. zero D. cannot be determined
| Situation | Choice |
|---|---|
| (a) Gravity on a ball while it rises after being thrown up | |
| (b) The normal force from the floor on a box sliding across a level floor | |
| (c) Kinetic friction on that same sliding box | |
| (d) String tension on a ball whirled in a horizontal circle at constant speed | |
| (e) Static friction on a crate riding (without slipping) in a truck bed as the truck speeds up | |
| (f) Gravity on a satellite in a circular orbit | |
| (g) A person's 50 N force on a box that moves 3.0 m (direction of the force not given) |
Explain the rule you used. What do all the "zero" situations have in common?
Back to topTwo carts start at rest on a frictionless track. Each is pushed by a constant 10 N horizontal force from the start line to the finish line, 2.0 m away. Cart 2 has twice the mass of Cart 1.
For each quantity, state whether it is greater for Cart 1, greater for Cart 2, or the same for both. If it's different, give the ratio.
| Quantity | Answer |
|---|---|
| (a) Work done by the 10 N force | |
| (b) Kinetic energy at the finish line | |
| (c) Speed at the finish line | |
| (d) Time to reach the finish line |
Explain each answer.
Back to topA 0.50 kg block is pushed against a spring ($k = 200\ \text{N/m}$), compressing it a distance $x$, and released. The block slides along a frictionless track and up the curved section.
The equation below is an energy conservation statement for some physical situation:
$$\tfrac12(400\ \text{N/m})(0.20\ \text{m})^2 = (0.50\ \text{kg})(10\ \text{m/s}^2)\,h \;+\; (0.30)(0.50\ \text{kg})(10\ \text{m/s}^2)(1.0\ \text{m})$$
A 1.0 kg cart is released from rest at point A on a frictionless track. The system is the cart and Earth, with $U_g = 0$ at point B.
A 60 kg student runs up a flight of stairs at constant speed in 6.0 s. The stairs rise 4.0 m vertically and are 10 m long measured along the slope. Find the student's average power output.
Student solution:
$F = mg = (60\ \text{kg})(10\ \text{m/s}^2) = 600\ \text{N}$
$W = Fd = (600\ \text{N})(10\ \text{m}) = 6000\ \text{J}$
$P = \dfrac{W}{t} = \dfrac{6000\ \text{J}}{6.0\ \text{s}} = 1000\ \text{W}$
Identify the error in the solution. Explain the physics of why it's an error, then find the correct power.
Back to topA 1000 kg car speeds up from 10 m/s to 20 m/s on a level road.
Jordan writes: $\Delta K = \tfrac12 m(\Delta v)^2 = \tfrac12(1000\ \text{kg})(10\ \text{m/s})^2 = 5.0\times10^4\ \text{J}$, so the engine must supply $5.0\times10^4\ \text{J}$ of energy to the car's motion.