# Finding acceleration of an object (with friction)

## Homework Statement

A 1,300-N crate is being pulled across a level floor at a constant speed by a force of 310 N at an angle of 20.0° above the horizontal. If coefficient of kinetic friction is 0.207, find the acceleration of the crate.

Fk= (uk)(N)
F=ma

## The Attempt at a Solution

Fx= 310cos20
Fy= 310sin20

N=1300-310sin20

Fk=(uk)(N) = (0.207)(1300-310sin20) = 247.15 N

sumFx= Fx-Fk =310cos20-247.15 = 44.15 N

F=ma=44.15
a=44.15/1300 =0.0340 m/s^2

Can someone point out what I did wrong? Thanks

Last edited:

SammyS
Staff Emeritus
Homework Helper
Gold Member
Doesn't gravity exert a force in the y direction, contributing to Fy ?

The Fy you have is downward. Do you want that to be negative ?

im not sure if I quite follow. Yes, force of gravity (1300N) is exerted in the y-direction:

F1y= 310sin20
N=1300-310sin20
sumFy=N+F1y-1300N=0

I'm not sure what is wrong. Can someone help me out?

Doc Al
Mentor
The applied force is downward:
a force of 310 N at an angle of 20.0° below the horizontal.

sorry that should be above the horizontal

Doc Al
Mentor
F=ma=44.15
a=44.15/1300 =0.0340 m/s^2
1300 N is the weight of the crate, not its mass.

so, how would i be able to find the acceleration?

Doc Al
Mentor
so, how would i be able to find the acceleration?
Just like you tried to do, only first find the mass of the crate. How do weight and mass relate to each other?

Got it, thanks. W=mg...find m using weight and use f=ma to find acceleration.