16 September, 11:20

# A stone is thrown with a speed v0 and returns to earth, as the drawing shows. Ignore friction and air resistance, and consider the initial and final locations of the stone. Which one of the following correctly describes the change ΔPE in the gravitational potential energy and the change ΔKE in the kinetic energy of the stone as it moves from its initial to its final location?A. ΔPE = 0 J and ΔKE = 0 JB. ΔPE is positive and ΔKE is negativeC. ΔPE = 0 J and ΔKE is positiveD. ΔPE is negative and ΔKE is positiveE. ΔPE = 0 J and ΔKE is negative

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1. 16 September, 13:20
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If the stone is thrown from the ground, the correct answer is A. If it is thrown from a height h, the correct answer is D.

Explanation:

Hi there!

I can't see the drawing but let's assume that initially, the stone is on the ground level. If that is the case, initially, the potential energy will be zero and when it returns to Earth it will also be zero. The potential energy depends on the height of the stone. If the final and initial height of the stone is zero, then the change in potential energy will also be zero:

ΔPE = final PE - initial PE

ΔPE = m · g · hf - m · g · hi (where hf and hi are the final and initial height respectively)

ΔPE = m · g (hf - hi)

ΔPE = m · g (0)

ΔPE = 0

Initially, the kinetic energy (KE) of the stone is the following:

KE = 1/2 · m · v0²

As the stone goes up, the kinetic energy is transformed into potential energy; but as the stone starts to fall, the acquired potential energy is transformed again into kinetic energy, so that the final and initial kinetic energy of the stone is the same.

Then:

ΔKE = final KE - initial KE = 0 (because final KE = initial KE).

Then, the correct answer is A.

Always ΔKE = - ΔPE due to the conservation of energy. Potential energy can't be acquired by the stone if there is no loss of kinetic energy and vice-versa.

Let's assume now that the stone is thrown from a height hi to the ground.

The final potential energy will be zero (becuase h = 0) but the initial PE will be:

PE = m · g · h1

Then:

ΔPE = final PE - initial PE = 0 - m · g · h1

Then ΔPE will be negative.

The initial kinetic energy will be:

KE = 1/2 · m · v0²

But the final kinetic energy will be equal to the initial kinetic energy plus the loss of potential energy (remember: if potential energy decreases, another type of energy has to increase, in this case, kinetic energy and vice-versa):

ΔKE = final KE - initial KE

ΔKE = 1/2 · m · v0² + m · g · h1 - 1/2 · m · v0²

ΔKE = m · g · h1

Then ΔKE will be positive and the correct answer would be D.