Mass on Inclined Plane with Friction

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Homework Help Overview

The discussion revolves around a block sliding down an inclined plane with friction, focusing on the calculations of acceleration, work done by gravitational force, energy expended overcoming friction, and kinetic energy. The problem involves concepts from dynamics and energy in physics.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the calculation of acceleration using forces acting on the block, including gravitational and frictional forces. There are questions about the work done by friction and how to calculate energy expended in overcoming this force. Some participants express confusion about the provided information and the distance involved in the calculations.

Discussion Status

Participants are actively engaging with the problem, clarifying details about the setup and calculations. There is an ongoing exploration of how to approach the work done by friction and its implications for energy calculations. Some guidance has been offered regarding the relationship between forces and work, particularly concerning friction.

Contextual Notes

There is a mention of a distance of 6 m that the block slides down the incline, which is relevant for calculating work done by friction. Participants are also addressing the potential lack of clarity in the problem statement regarding the necessary parameters for calculations.

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

 
Last edited:
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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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