Work and kinetic energy

  • Thread starter shaunanana
  • Start date
  • #1

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?


Homework Equations



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
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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Welcome to PF!

Hi shaunanana! Welcome to PF! :smile:

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

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 = … ? :smile:
 
  • #3
Oh wow, that's so simple...maybe I should stop trying to do my homework so late at night!

Thank you!!! :)
 

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