# Work done by friction on a conveyor belt [closed]

At a factory, a 300-kg crate is dropped vertically from a packing machine onto a conveyor belt (I apologize for not having a picture) moving at a speed of 1.2 m/s. The coefficient of kinetic friction between the belt and the crate is .40. After a short time, slipping between the belt and the crate ceases and the crate then moves along the belt. For the period of time during which the crate is being brought to rest relative to the belt, calculate, for a coordinate system at rest in the factory, (A) the work done by friction...

I can do the entirety of the problem except I am unsure of this first part and need a little help from the community.

I tried:

W(friction) = E = (delta)KE = 1/2m(vf)^2 - 1/2m(vi)^2 = 216J.


This answer is correct however I am uneasy with the approach to the solution, if it is correct(doubtful) could someone explain how? otherwise a solution would be appreciated. -Thank you!

## closed as off-topic by user10851, Kyle Kanos, ACuriousMind♦, John Rennie, David Z♦Mar 28 '16 at 15:42

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Community, Kyle Kanos, ACuriousMind, John Rennie, David Z
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They expected you to solve this question by using the definition of work rather than using the conservation of mechanical energy. Therefore, if we consider the definition of work; $$W=\int Fdx=F_fx$$ where, $F_f$ is the force due to the friction on the object which can be calculated as $300g\times0.4=120g$. Moreover, to find the displacement of the object the acceleration due to the frictional force and the time passed before it comes to rest relative to the conveyor belt, which are $0.4g ms^{-2}$ and $\frac 3g s$ respectively, should be found. Therefore, the displacement is be $\frac {1.8}g m$. Thus, the work done by friction is; $$W=\frac{1.8}g\times 120g=216 J$$ However, as I said this is a longer and more cumbersome approach to this problem and both your and this solution are equally correct.