Motion & Energy: Calculate Total Energy & Friction Coefficient

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The discussion revolves around a physics problem involving motion and energy, specifically calculating total energy expended and the coefficient of friction for a mass acted upon by a horizontal force. The scenario includes a mass of 6 kg moving a distance of 5 m under the influence of an 80 N force, with considerations for acceleration and friction.

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  • Mixed

Approaches and Questions Raised

  • Participants explore the calculation of total energy expended, questioning whether it should include energy lost to friction or only the energy required for acceleration. There are discussions about the relationship between work done by the force and the energy gained by the mass.

Discussion Status

Multiple interpretations of the problem are being explored, particularly regarding the definition of total energy expended. Some participants suggest providing both interpretations in their answers, while others emphasize the importance of considering friction in the calculations.

Contextual Notes

There is uncertainty about the wording of the question and whether it implies the inclusion of frictional energy in the total energy calculation. Participants express concern about the implications of the problem's phrasing on the complexity of the solution.

jimmy4
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I have attempted this question, please feel feel to comment.

A horizontal force of 80 N acts on a mass of 6 kg resting on a horizontal surface. The mass is initially at rest and covers a distance of 5m in 0.92 s under the action of the force. Assuming that there is no energy lost to air resistance and therefore that the acceleration is constant.
1. calculate the total energy expended in the acceleration
2. calculate the coefficient of friction between the mass and the surface.

Ans attempt.

1. acceleration s = (1/2) at^2
s = (1/2) a(0.92)^2
5 = 0.4232a
a = 5/0.4232
a = 11.815 m/s

Final Velosity v = a * t.
v = 11.815 * 0.92 = 10.87ms^2
Initial Velosity u = 11.815 * 0 = 0

Kinetic Energy Ke = 1/2 (mv^2)
= (1/2) 6 * 10.87^2
= 354.46 Joules

2. Force = m * a the fifferance between 80 N and the friction.
6 * 11.815 = 70.89 N

P = 80 - 70.79 = 9.11 N

P = (mu)Rn
mu = P/Rn
9.11/58.86
mu = 0.154

Any advice would be welcome. Jimmy
 
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Isn't there also an energy lost due to friction that adds to the total?

Hint: How much total work was done by the 80N force on the block? How does that differ from the energy gained by the block? (You computed this above.) Can you relate this discrepancy to the coefficient of friction?
 
G01. ... I think the question is slightly missleading. The total energy expended is just force * distance but the question asks..

1. calculate the total energy expended in the acceleration

Is that asking for the total energy required to accelerate the object OR just that fraction required to accelerate it and ignoring the fraction required to overcome friction?

I'd give both answers and let the examiner sort it out :-)
 
CWatters said:
G01. ... I think the question is slightly missleading. The total energy expended is just force * distance but the question asks..



Is that asking for the total energy required to accelerate the object OR just that fraction required to accelerate it and ignoring the fraction required to overcome friction?

I'd give both answers and let the examiner sort it out :-)

I pretty sure it's asking for the total energy expended. Overcoming friction is an important part of the acceleration process. :)
 
G01 said:
I pretty sure it's asking for the total energy expended. Overcoming friction is an important part of the acceleration process. :)

I think you are right, for a body to accelerate it would have to expend energy to overcome friction.

The answer "354.47 J " would have been correct if the question explicitly asked about the amount used to increase the Kinetic Energy .
 
I was concerned that makes the question rather easy. Just 80N x 5m = 400 Joules.
 
CWatters said:
I was concerned that makes the question rather easy. Just 80N x 5m = 400 Joules.

Yes, part 1 is easy if the question is interpreted this way. But all the work you did above is still important for part 2.
 

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