Mass on Inclined Plane with Friction

AI Thread Summary
A block of mass 2.5 kg slides down a 31° incline with a coefficient of kinetic friction of 0.2, resulting in an acceleration of 3.37 m/s². The work done by gravitational force on the block is calculated to be approximately 75.79 J. The discussion focuses on determining the energy expended to overcome friction and the kinetic energy of the block after sliding 6 m down the incline. The normal force does no work on the block, as it acts perpendicular to the motion. The participants clarify the calculations for friction and its contribution to work, emphasizing the importance of understanding forces acting over distance.
r34racer01
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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?

Homework Statement


Homework Equations


The Attempt at a Solution

 
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r34racer01 said:
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

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?

Homework Statement



Homework Equations



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.
 
LowlyPion said:
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.
 
r34racer01 said:
Ah, no I've given everything.

Ok then what is the distance?
 
LowlyPion said:
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.
 
r34racer01 said:
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.
 
LowlyPion said:
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.
 
r34racer01 said:
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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