How do you find potential and kinetic energy? I've gotten these questions wrong on my test and I do not understand why.

a 3-kg object is thrown horizontally off a 12-m high cliff. The Kinetic energy of the object just before it hits the ground is 500 J. ignore air resistance.

a) what is the potential energy of the object just after it is thrown? Set gravitational potential energy equal to zero when the object hits the ground

my answer:kinetic energy just before the object hits the ground is 500 J, which means that potential energy would be 500 J right after it is thrown

b) What is the kinetic energy of the object just after it is thrown?

my answer: kinetic energy would be 0 J because the kinetic energy was 500 J just before it hit the ground


3 Answers 3


First you need to understand when the conservation of energy works.

It works only when the forces acting on the body are conservative i.e, forces for which the net work done in a closed loop is zero.

When the object is at the cliff, it has a certain Potential energy which is equal to $mgh$. Energy for the object will be conserved after the event of throwing the object because after this event there are only conservative forces acting on the object.

Now when you throw the object, you actually apply external force, which gives it a certain velocity. The object will gain the velocity the instant after the throwing event completes. Therefore, just after throwing the KE of the object won't be zero but it will be equal to $\frac{1}{2}mv^2$, where $v$ the initial velocity of the object in horizontal direction.

Now when the object comes down, the PE converts into KE. Gravity did -ve work when the object was taken at the cliff before it was thrown, and the gravity will do equal amount of +ve work when the object comes down.

Summing up, the KE of the object at the bottom will be equal to the initial kinetic energy and the initial potential energy.

  • $\begingroup$ how would I find velocity to figure out the KE? $\endgroup$
    – Callie
    May 4, 2017 at 19:29
  • $\begingroup$ Hint : You have the final KE. $\endgroup$
    – Mitchell
    May 5, 2017 at 3:02

The final 500J is the sum of the initial KE due to the horizontal throw and the gain in KE from the change in potential energy. The initial PE is $mg\times$ the 12 meters. From this you can find the initial horizontal velocity if you need it.


It is a bit of trick question - the key word is "thrown"

Potential Energy In your answer you start at the ground and then work backwards. To check it, work forwards and see if you come up with something different. When 'thrown' off the cliff, what information do you have? Only the height (12m). So look at your gravitation potential energy mass x g x height. 3kg x g x 12. Let's say g is 10 because I am lazy.

PE = 3*10*12 = 360J

Hmm, this seems too low. How did the item go from 360J potential energy to 500j kinetic energy? Your gut is correct, it can't. So, we re-read the question and see that it was thrown. That means it must have had some kinetic energy even before it starting dropping due to gravity.

Kinetic Energy Well, in the above you already saw that 360j of potential energy was released by dropping 12meters. So to work out the extra energy we just do 500 minus 360 = 140J

Energy is conserved here - but from the moment that the object leaves the person's hands. THEY put 140J of energy into their throw.

If they dropped the item, your answer would be correct. It would start with zero speed, zero KE, and then gain KE as it falls. But it would reach only 360J.

I hope that helps.


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