# Mechanics, energy, acceleration.

1. Feb 15, 2012

### oxon88

1. The problem statement, all variables and given/known data

A horizontal force of 80N acts on a mass of 6Kg resting on a horizontal surface. The mass is initially at rest and covers a distance of 5m in 0.92s under the action of the force. Assuming there are no energy losses due to air resistance and therefore that the acceleration is constant

i) calculate the total energy expended in the acceleration

ii) calculate the coefficient of friction between the mass and the surface.

2. Relevant equations

v = s/t

a = (v-u) / t

s = ut + (1/2)a.t2

F = m.a

Ke = 1/2*m*V^2

3. The attempt at a solution

first I calculated the acceleration:

s = (1/2).a.t2

5 = (1/2)a.0.922

10/0.922 = a = 11.815 ms-2

Then i calculated the net force:

F = m.a

F = 6 * 11.815 = 70.89N

therefore i think friction = 80N - 70.89N = 9.11N Would this be the coefficient of friction?

Next i calculated the speed after 0.92s. s = d/t = 5/0.92 = 5.435 ms-1

Then calculate kinetic energy: = (1/2)m.v2 = 3*5.4352 = 88.62J

so now would the total energy expended = kinetic energy + Friction force * 5m?

= 88.62 + 9.11 * 5 = 488.65 Nm

can anyone verify this? and let me know if i'm way off please?

2. Feb 15, 2012

### PeterO

This line [red] of your calculation calculates the average speed only. This object is accelerating!!

EDIT: And the line before, you calculated the actual friction; not the co-efficient of friction.

Last edited: Feb 15, 2012
3. Feb 15, 2012

### oxon88

ok so because its accelerating, would I use s = (U + V * t) / 2

s = (5.435 *0.92) / 2 = 2.5 ms-1 ??

4. Feb 15, 2012

### oxon88

is the acceleration not constant though? so surely speed will be constant?

5. Feb 15, 2012

### PeterO

That formula is poorly written, and not appropriate.

This object was accelerating at 11+ m/s/s - if your calculations are correct, and maintaind that acceleration for almost a whole second, so will have reached a speed approaching 11 m/s.

EDIT: that formula should have been written s = (U+V)*t/2

Last edited: Feb 15, 2012
6. Feb 15, 2012

### PeterO

Speed is constant when the acceleration is zero!!!

Constant acceleration means the speed continually increases, at a steady rate.

eg: an acceleration of 10m/s2 means the speed increases from 0 -10 in one second, then up to 20 m/s [another 10] in the next second, and yet another 10 to 30 m/s after 3 seconds.

That's what acceleration means.

7. Feb 17, 2012

### oxon88

ok i had a re-think.

can i not just use:

Work done = Force x Distance

= 80N x 5m = 400J

Power = Work Done / Time

= 400j / 0.92s = 434.783 Watts

8. Feb 17, 2012

### PeterO

BUT

i) calculate the total energy expended in the acceleration

ii) calculate the coefficient of friction between the mass and the surface.

The original question(s) didn't ask anything about Power ??

9. Feb 17, 2012

### oxon88

ok am i any closer here?....

s = (a.t2)/2

5 = a.(0.922)/2

5 = a.(0.4232)

a = 5/(0.4232) = 11.815 ms-1

so if the acceleration is 11.815 ms-1 can i then use v = a.t?

V5 = 11.815*0.92 = 10.87 ms2

v0 = 11.815*0 = 0 ms2

then do i work out the Ke?

Ke = (1/2)m.v2 = (1/2)80x10.872 = 4726.276 J

would this give me the total energy expended?

10. Feb 17, 2012

### PeterO

Getting closer. when you calculated the Ke, you used 80 for the mass. 80 was the force.

The energy expended will be greater than that, since work was done against friction. That's where part (ii) comes in.

11. Feb 20, 2012

### oxon88

whoops. ok so if i use the mass this time....

Ke = (1/2) 6000 x 10.872

=354470.7 j

will i need to add this to the energy expended due to friction?

any advice on working out the coefficient of friction?

12. Feb 20, 2012

### PeterO

Sorry, but the mass is 6 not 6000.

base units in Physics are metres, kg and seconds. I first learnt it as the mks system, but better known as SI units [International system of units - but in French where the word for System comes before the word for International]

13. Feb 21, 2012

### oxon88

ok

Ke = (1/2) 6 x 10.872

= 354.47 j

what else is needed to find the total energy expended?

Last edited: Feb 21, 2012
14. Feb 21, 2012

### PeterO

re-visiting the original post:

i) calculate the total energy expended in the acceleration

ii) calculate the coefficient of friction between the mass and the surface.

(i) I find a little unclear. I could put a case for the answer here being what was requested, and then again the Work done [Your post #7 before you went on pursuing Power] may be what is sought.

Certainly you do a lot of work, and much of that is realised by acceleration to the speed we reached.
The "lost" energy [ie Work - final Ek} will enable you to find the work done against friction - and thus the size of the frictional force. From that you can calculate the co-efficient of friction; μ

15. Feb 22, 2012

### oxon88

Yes i see your point.

So i could answer the question by saying that the total energy expended is 400 J (Work = F x D)

Then I could go on further and say that the energy expended in the acceleration is 354.47 J, therefore the energy expended because of friction is 45.53 J

can you offer any advice on finding the coefficient of friction?

i have had a go and come up with this:

F=μRN

RN=m.g = 6*9.81 = 58.86N

Can i now use the friction I calulated previously (45.53J) in the formula?

45.53 = μ.58.86

μ = 45.53 / 58.86 = 0.774

16. Feb 22, 2012

### PeterO

Not quite.

Watch you quantities F is a force -measured in Newtons

That 45.53 was 45.53 Joules - a measure of work or energy.

You need to find what distance was covered, so that you can find the size of the force.
OR
Tou could find out the acceleration [I think you already have] and work out what net force is necessary for that - lets say it was 78 N [I just made up that figure]. That would mean that friction must have been 2N, to reduce the effect of the 80N applied force to a mere 78N

Using the correct derived Force, you will get the correct coefficient of friction

If you are worried about change of method on the way through this, you have to become familiar with the fact that these problems can be worked out in a variety of ways - and a kinematics approach may be useful for some values, work/energy for others, impulse/momentum for still others.

Each method can be a good starting point, but usually one method is the quickest and easiest depending what you are trying to calculate.

Note: A few times here you have strayed onto the wrong quantity - and the easiest way to keep track is follow the units.

17. Feb 23, 2012

### oxon88

Ok right so if i say FNet=m.a

FNet=6*11.815 = 70.89

then i take that away from the 80N (80N - 70.89 = 9.11N

and then use F = μ.R

9.11 = μ(58.86)

9.11/58.86 = μ

μ = 0.155

so the materials would likely be metal on metal?

Does that look better? Would i not need to take gravity into account when working out FNet? (somthing like: FNet= m.a.g)??

18. Feb 23, 2012

### PeterO

You used gravity when you calculated the Normal reaction force - 58.86N

As for the net force - you don't need gravity, just F[sub[net[/sub] = ma

19. Feb 23, 2012

### oxon88

ok that makes sense. so the answers of μ = 0.155 would be correct?

20. Feb 23, 2012

### PeterO

Looks good. - and is a reasonable answer.