Multistep stumper

1. Oct 31, 2006

Sportsman4920

A 5.4 kg block is pushed 3.0 m up a rough 37° inclined plane by a horizontal force of 75 N. If the initial speed of the block is 2.2 m/s up the plane and a constant kinetic friction force of 25 N opposes the motion, *calculate the following.
(a) the initial kinetic energy of the block
J
(b) the work done by the 75 N force
J
(c) the work done by the friction force
J
(d) the work done by gravity
J
(e) the work done by the normal force
J
(f) the final kinetic energy of the block
J

2. Oct 31, 2006

stunner5000pt

what have u done to answer them yourself?

3. Oct 31, 2006

BishopUser

$$W = |F||D|cos\theta$$ theta being the angle between force and displacement vectors. That should take care of the first 5. The last one looks like $$W_{net} = \Delta KE$$

Last edited: Oct 31, 2006
4. Oct 31, 2006

Sportsman4920

thank you

5. Oct 31, 2006

bosox3790

I tried to use the formula but I cant get the right answer.

6. Oct 31, 2006

bosox3790

how would you do part be on this question?

7. Oct 31, 2006

BishopUser

In part B the force is 75N the distance is 3m and the angle between them is 37 degrees, so using the formula the work should be 180 N*m

8. Nov 4, 2006

Sportsman4920

Thanks, now i'm trying to find the work done by friction. so I did W=25N*5.2(distance)*cos(theta but it didn't work