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A bullet with mass $8g$ flying at the speed of $v_0=600$ $m/s$ strikes an apple and comes out at the speed of $500$ $m/s$. Calculate the work ($A$) of the force exerted by the apple on the bullet.

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The answer given in my book is $A=\dfrac{m}{2}(v^2-v_0^2)=-440$ $J$. That is equal to $A=\dfrac{mv^2}{2}-\dfrac{mv_0^2}{2}=E_k-E{k_0}.$ Why don't they take account of the potential energy? According to the law of conservation of energy: $A=E-E_0$ where $E$ and $E_0$ are mechanicals energies (the sum of potential energy and kinetic energy). I think that $E=\dfrac{mv^2}{2}+mgh$ and $E=\dfrac{mv_0^2}{2}+mgh$, but I am not sure if $h$ changes. Thank you in advance!

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    $\begingroup$ The height of the bullet above a horizontal line does not change so there is no change in gravitational potential energy. $\endgroup$
    – Farcher
    Nov 23, 2019 at 11:11
  • $\begingroup$ @Farcher That's an answer $\endgroup$
    – FGSUZ
    Nov 23, 2019 at 11:20
  • $\begingroup$ @Farcher, so we should take account of the potential energy, but it's just equal in this problem, right? $\endgroup$ Nov 23, 2019 at 12:16

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Assuming that the apple has a diameter $d$ of 5 cm, the bullet takes between $d/v = 83\ \mu s$ and $d/{v_0} = 100\ \mu s$ to cross it. Assuming further that the bullet enters the apple pretty much horizontally, the drop in height due to gravity during the traversal is of the order of $\Delta h \approx g t^2 \sim 100\ \text{nm}$. The variation of gravitational energy is then of the order of $mgh \sim 10\ \text{nJ} \ll 440\ \text J$.

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    $\begingroup$ It might be better to show that any increase in speed, and therefore kinetic energy, due to gravity is negligible since any change in gravitational potential energy is due to work done by gravity not the apple. $\endgroup$
    – Bob D
    Nov 23, 2019 at 13:30
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You are right to wonder about the possibility of a change in gravitational potential energy. Actually, the problem is poorly worded.

The problem appears to apply the work energy theorem which states that the net work done on an object equals its change in kinetic energy. Note my emphasis on "net". Thats because there are two external forces acting on the bullet that can do work, the force by the apple and the force of gravity. The problem only asks for the work done by the apple.

Although @Phoenix87 has shown that as a practical matter the effect of gravity would be negligible the problem should technically state the bullet enters and exits the apple at the same level or to equivalently state that the change in speed due to gravity is negligible. Then you would know the change in kinetic energy is due only to the work by the apple. Otherwise, there can be an increase in kinetic energy (increase in downward component of speed) due to positive work done by gravity.

Alternatively to stipulating negligible speed change due to gravity or no change in height, the problem can simply ask what is the net work done on the bullet, which would mean the work done by the apple plus gravity. Then the equation applies without ambiguity.

Hope this helps.

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