Calculating Work Done by Force on a Block

[lauriecherie]

Homework Statement

A 100 kg block is pulled at a constant speed of 3 m/s across a horizontal floor by an applied foce of 145 N directed 37 degrees above the horizontal. What is the rate at which the force does work on the block? _____ W

Homework Equations

work = force (dot product) displacement
work = .5m * v^2 (final) - .5m *v^2 (initial)

The Attempt at a Solution

i used the second equation and plugged in my mass as 100kg and my final velocity as 3. so i got .5(100)(3^2) = 450 W. That seems like an awful lot to me ?

 

[sArGe99]

Its the rate at which force does work.
What is the rate at which work is done known as?

 

[lauriecherie]

Is it power?

 

[lauriecherie]

This is something else I've tried.

Power = work / time = force (dot product) displacement = force * average velocity * cos(37 desgrees). I cam out with 347. 41 W. Am I doing this correctly?

 

[lauriecherie]

Solved. Thanks for jogging my memory!