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## Homework Statement

A 25-kg box slides, from rest, down a 9.0-m-long incline that makes an angle of

15° with the horizontal. The speed of the box when it reaches the bottom of the

incline is 2.4 m/s. (a) What is the coefficient of kinetic friction between the box

and the surface of the incline? (b) How much work is done on the box by the

force of friction? (c) What is the change in the potential energy of the box?

## Homework Equations

F

_{x}= mg * sin15

F

_{n}= mg * cos15

F

_{fr}= u

_{k}* Fn = u

_{k}* mg * cos15

Vf

^{2}-Vo

^{2}= 2 *a*d

## The Attempt at a Solution

Having an issue with kinetic friction calculation here

Taking the force components and applying Newton's 2nd law -> ƩF = ma

F

_{x}- F

_{fr }= ma

_{x}

mg * sin 15 - u

_{k}(mg * cos 15) = m * a

_{x}

Since I want uk and the final velocity is given, I calculate accel -> Vf

^{2}- Vo

^{2}= 2*a*d so a = 2.4m/s / (2*9.0m) = 0.32 m/s^2

Can I put it into mg * sin 15 - u

_{k}(mg * cos 15) = m * a

_{x}and end up with u

_{k}= 0.2345 ?

I can't get over that a

_{x}isn't constant acceleration and I don't think I can use it in this case....

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