# Homework Help: Work and energy problem-need help

1. Mar 19, 2006

### chazgurl4life

A 4.6 kg block is pushed 5.6 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 inital KE of the block
B) the work done by the 75N force
C) The work done by friction force
D) the work done bt gravity.

Now i figured out that a) initial KE = 1/2mv^2=.5(4.6kg)(cos37 degrees)
B) Work done by the 75N= Fdcostheta=75N(5.6m)(cose37)
Now c &d are presenting a problem for me

2. Mar 19, 2006

### Hootenanny

Staff Emeritus
(a)Your initial kinetic energy is wrong, you don't need to resolve because the velocity is given up the plane.

(c)is simply the energy dissappated due to friction. So it is the frictional force mulitplied by distance

3. Mar 19, 2006

### chazgurl4life

actually my initial KE was correct according to my notes but thank you very much..i appreciate it

4. Mar 19, 2006

### chazgurl4life

should the wortk done by friction force supposed to be negative because its opposite to going up the plane

if everything works out should work= 25 N (-5.6m)=-140 Joules

5. Mar 19, 2006

### Hootenanny

Staff Emeritus
Well, I'm afraid your notes are wrong. Either that or you've typed the question wrong.

6. Mar 19, 2006

### siddharth

Nope, your calculation of initial KE isn't right. I think you should try understanding the concepts, rather than try applying equations from your notes.

7. Mar 19, 2006

### chazgurl4life

lol...im sorry made a typo above ...sorry PE= .1/2mv^2 = so its supposed to be Pe=.5(4.6 kg)(2.2 m/s)^2=>11.132 Joules

8. Mar 19, 2006

### Hootenanny

Staff Emeritus
Yep, that's right.

9. Mar 19, 2006

### chazgurl4life

10. Mar 19, 2006

### chazgurl4life

How do i figure out the final KE and the work done by normal force?

11. Mar 19, 2006

### Libertine

The total final KE must be the same as the total starting KE, so:
Starting KE - Work Done by block + Work done on block = Final KE + GPE

12. Mar 19, 2006

### Hootenanny

Staff Emeritus
Work done by the normal force is not the same as work done by gravity. I believe the question asks for the work done by gravity.

13. Mar 19, 2006

### chazgurl4life

so if i have this correct the work done on the block is performed by gravity & fiction and the work done by the block is the 75 N force:
So :
111.132 J -335 J(75cos37degrees) + -140 J( friction)+ Work done by gravity = Final Ke
is this correct

14. Mar 19, 2006

### Hootenanny

Staff Emeritus

$$11.132 + 5.6\times75\cos 37 - 140J + W_{gravity} = E_{final}$$

You are giving the block energy when you apply the force, therefore it must be positive.

15. Mar 19, 2006

### chazgurl4life

is the work done by gravity attributed to mass(distance) * 9.8 or must i include the incline which would make it mgcostheta=Work done by gravity

16. Mar 19, 2006

### Hootenanny

Staff Emeritus
Work done by gravity is simply mgh - Force (mg) times distance moved (h)

17. Mar 19, 2006

### chazgurl4life

hello is anyone there