Vada's wagon rolls on a level sidewalk, carrying 30 J of kinetic energy. Over the next stretch, the force from her hand on the handle does +40 J of work on the wagon, while friction does −15 J.
Two forces, two signed works. What happens to the wagon's kinetic energy?
First add the signed works, the move you met earlier in this topic: +40 J and −15 J give a net work of 40 − 15 = 25, so W_net = +25 J.
Here is the theorem that spends that number. The net work done on an object equals its change in kinetic energy.
For the wagon: ΔKE = +25 J, so its kinetic energy climbs from 30 J to 30 + 25 = 55, reaching 55 J.
The trade is exact, joule for joule: +50 J of net work raises an object's KE by exactly 50 J.
A net work of +26 J is done on a rolling ball. What happens to the ball's kinetic energy?
A 2.0 kg cart carries 12 J of kinetic energy. A rope's force then does +28 J of work on it while friction does −10 J. Calculate the cart's new kinetic energy.
Write down the values:
KE at the start = 12 J; works done: +28 J by the rope's force, −10 J by friction
Net work: 28 − 10 = 18, so W_net = +18 J.
W_net = ΔKE, so the kinetic energy climbs by 18 J.
Add the change:
new KE = starting KE + ΔKE
new KE = 12 + 18 = 30
The cart's new kinetic energy is 30 J.
A skateboard carries 60 J of kinetic energy. Over the next stretch, the net work done on it is +24 J. Enter its new kinetic energy, in J.
Negative net work takes the same road out. W_net = −20 J means the kinetic energy drops by exactly 20 J.
A sled crosses a gravel patch carrying 72 J of kinetic energy, and friction does −50 J of work on it as it slides. No other force does work on it. Calculate its kinetic energy after the patch.
Write down the values:
KE at the start = 72 J; net work = −50 J
W_net = ΔKE, so the kinetic energy drops by 50 J.
Subtract the change:
new KE = starting KE + ΔKE
new KE = 72 − 50 = 22
The sled's kinetic energy is 22 J after the patch.
The net work done on a gliding canoe over a stretch is −18 J. What happens to its kinetic energy?
The theorem adds no new quantity. Net work and KE = ½mv² were both taught earlier in this topic; the theorem says the two move together.
Why the net work lands exactly in the kinetic energy is the next lesson's question.
A delivery robot carries 150 J of kinetic energy. Braking, the net work done on it is −45 J. Enter its kinetic energy afterward, in J.
A toboggan carrying 10 J of kinetic energy slides down a long slope. Over the slide, gravity does +45 J of work on it and friction does −20 J. Enter its kinetic energy at the bottom, in J.
Which statement is the work-energy theorem?