Mass on Inclined Plane with Friction


by r34racer01
Tags: friction, inclined, mass, plane
r34racer01
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#1
Oct12-08, 02:32 PM
P: 63
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 ľk = 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
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LowlyPion
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#2
Oct12-08, 03:38 PM
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Quote Quote by r34racer01 View Post
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 ľk = 0.2.

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

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
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.
r34racer01
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#3
Oct12-08, 07:33 PM
P: 63
Quote Quote by LowlyPion View Post
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.
Ah, no I've given everything.

LowlyPion
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#4
Oct12-08, 08:29 PM
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Mass on Inclined Plane with Friction


Quote Quote by r34racer01 View Post
Ah, no I've given everything.
Ok then what is the distance?
r34racer01
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#5
Oct12-08, 08:36 PM
P: 63
Quote Quote by LowlyPion View Post
Ok then what is the distance?
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."

Sorry about that.
LowlyPion
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#6
Oct12-08, 09:45 PM
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Quote Quote by r34racer01 View Post
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."

Sorry about that.
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.
r34racer01
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#7
Oct12-08, 11:46 PM
P: 63
Quote Quote by LowlyPion View Post
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.
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.
LowlyPion
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#8
Oct12-08, 11:51 PM
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Quote Quote by r34racer01 View Post
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.
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.


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