Motion & Energy: Calculate Total Energy & Friction Coefficient

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A horizontal force of 80 N acts on a 6 kg mass, which accelerates over 5 meters in 0.92 seconds, leading to a calculated acceleration of 11.815 m/s². The kinetic energy gained by the mass is approximately 354.46 Joules, but the total energy expended, considering the force and distance, is 400 Joules. The discrepancy between these values highlights the energy lost to friction, which is estimated using the difference in force calculations. The coefficient of friction is calculated to be approximately 0.154, indicating the friction's role in the energy dynamics. Clarification on whether the total energy expended includes friction is debated, with suggestions to provide both interpretations in responses.
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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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