# Work and kinetic energy

## Homework Statement

a 300 kg crate is on a rough surface inclined at 30 degrees. A constant force P=2400 N is applied horizontally to the crate. The force pushes the crate a distance of 3.0 m up the incline, in a time interval of 9.0 s, and the velocity changes from v1=0.7 m/s and v2=2.9 m/s. What is the work done by the weight?

WD=F.d

## The Attempt at a Solution

i have found the parallel components:
Fnet=Psin(theta) -mgsin(theta)
Psin(theta)-mgsin(theta)=ma
and the perpendicular components:
Fnet=Fn-mg
Fn=mg
I'm not 100% sure any of this is correct, and now I don't know where to go with it

tiny-tim
Homework Helper
Welcome to PF!

Hi shaunanana! Welcome to PF!

(have a theta: θ and try using the X2 tag just above the Reply box )

You're only asked for the work done by the weight (the mg force), so most of the information in the question is unnecessary.

And work done = force "dot" distance.

So work done by the weight = … ?

Oh wow, that's so simple...maybe I should stop trying to do my homework so late at night!

Thank you!!! :)