# Pushing a block up a hill (kinematics/dynamics)

## Homework Statement

A small box is at the bottom of a ramp tilted at an angle of 40 degrees above horizontal. The box is given a push and it then slides up the ramp 2.0 seconds before sliding back down. The coefficient of sliding friction is 0.15. Find the time for it to slide back to the point it was released.

Fnet = ma

## The Attempt at a Solution

a = - gsin$$\theta$$ - ugcos$$\theta$$
a = - 6.41 m/s2

Up:

v2 = 0
t = 2s
a = - 6.41 m/s2

d = -(-1/2at2)
d = 12.83 m

Down:

v1 = 0
a = gsin$$\theta$$ - ugcos$$\theta$$ = 6.18 m/s2
d = 12.83 m

t = sqrt(d/(1/2a)) = 2.04 s

The answer is supposed to be 2.4 s.

What'd I do wrong?

## Answers and Replies

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kuruman
Homework Helper
Gold Member
Recheck your calculation for the acceleration down. It is awfully close to the acceleration up.

*** On edit ***
Actually, your value for acceleration up the incline is incorrect which gives you an incorrect value for the distance traveled. Redo your calculation and be sure you set your calculator to "degrees".

Last edited:
Recheck your calculation for the acceleration down. It is awfully close to the acceleration up.

*** On edit ***
Actually, your value for acceleration up the incline is incorrect which gives you an incorrect value for the distance traveled. Redo your calculation and be sure you set your calculator to "degrees".
I see what I did wrong, on the paper I was doing the problem on, I wrote a = -gsin(theta) - ucos(theta), and missed the g on the second term, so I kept plugging that into my calculator. For some reason I typed it up correctly here :s

Thanks!