# Mass on Inclined Plane with Friction

1. Oct 12, 2008

### r34racer01

A block of mass M = 2.5 kg is released from rest and slides down an incline that makes an angle q = 31° with the horizontal. The coefficient of kinetic friction between the block and the incline is µk = 0.2.

a) What is the acceleration of the block down the inclined plane?
I got a = 3.37

The next four questions are concerned with what has happened after the block slides a distance d = 6 m down the plane.

b) How much work was done on the block by the Earth's gravitational force?
I got Wg = 75.7878

c) How much energy was expended in overcoming the frictional force (thus producing heat, etc.)?

d) What is the kinetic energy of the block?

e) The plane exerts a normal force (perpendicular to its surface) on the block. How much work was done on the block by this normal force?
Easy that's 0.

I just can't get how to get E-heat or KE?

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited: Oct 12, 2008
2. Oct 12, 2008

### LowlyPion

Let me guess there is a distance given in a figure that you haven't provided? Work is N-m and I'm not seeing any meters.

3. Oct 12, 2008

### r34racer01

Ah, no I've given everything.

4. Oct 12, 2008

### LowlyPion

Ok then what is the distance?

5. Oct 12, 2008

### r34racer01

Oh my bad I didn't see that after pt a they say

"The next four questions are concerned with what has happened after the block slides a distance d = 6 m down the plane."

6. Oct 12, 2008

### LowlyPion

OK then what was the term that you used for determining the friction in calculating the acceleration? It is that force acting over the distance that will be the work done by friction.

7. Oct 12, 2008

### r34racer01

Well what I did was F = ma, in this case (mg sin 31) - Ff = ma.
So ((2.5*9.81) sin 31) - (0.2(2.5*9.81 cos 31) = 2.5*a = 3.37
To determine friction I just calculated the normal force * uk. In terms of forces they act over the distance but not in terms of work at least for normal force.

8. Oct 12, 2008

### LowlyPion

Friction does act over that distance however. While its scalar magnitude is a function of the normal force its direction of action is parallel to the incline. Hence the frictional force component times distance does have a non zero contribution to work - albeit negative.