# Motion and Energy

1. Jan 15, 2013

### jimmy4

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

2. Jan 15, 2013

### G01

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?

3. Jan 16, 2013

### CWatters

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 :-)

4. Jan 16, 2013

### G01

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

5. Jan 16, 2013

### hms.tech

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 .

6. Jan 16, 2013

### CWatters

I was concerned that makes the question rather easy. Just 80N x 5m = 400 Joules.

7. Jan 16, 2013

### G01

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.