# Work on a stalled car

1. Nov 5, 2012

### KnockKnock

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

Three people, each of whom pushes with a force of 130N, can just keep a stalled car (mass 1400kg) moving along a level road. They each push with a force of 140N to get it to a service station 100m away.

2. Relevant equations

a) How much work do they do in moving the car to the service station?
b) What is the kinetic energy of the automobile on arriving?
c) What is the automobile's speed on arrival?

3. The attempt at a solution

a) I said that: $Work=[(140N \times 3) - (130N \times 3)] \times 100m=3000J$, but I'm unsure if I correctly calculated the average force. My thinking was that the car was just moving with 390N of force (130N multiplied by 3 people), so it had no net force and was in constant motion. Once the extra 10N per person of force were applied, the car moved at an accelerated rate, so the net force was 420N-390N=30N. I can't figure out if I'm overthinking this or not.

b) I can't figure out how to get kinetic energy without velocity $(KE= \frac{1}{2}mv^2)$. I can find the CHANGE in kinetic energy (work), but without knowing how much KE the car had before the 100m stretch of road I can't figure it out.

c) If I can get KE, I can easily solve for v.

Last edited: Nov 5, 2012
2. Nov 5, 2012

### KnockKnock

UPDATE: I asked one of my friends, and he claims that the work done in part a (which he thinks I did right) is ALSO equal to the kinetic energy in part b. I'm not sure if that's true because the car IS moving, so it has some KE even before it is pushed with the additional force over the 100m road.

Am I correct in assuming the total KE has to be greater than just 3000J, or would the motion of the car before the 100m not be considered a factor in this problem?

3. Nov 6, 2012

### haruspex

According to your analysis for (a), if they had just kept the car moving (130N each) they would have done no work. I'm sure they would have felt that they had worked quite hard.
For (b), how much work had they done, and how much of that had gone into overcoming friction? Where has the rest gone?

4. Nov 7, 2012

### KnockKnock

But why wouldn't there be work done when the 130N are applied? Isn't it still a force being applied over a distance, and thus work?

My understanding of work is, more or less, a change in kinetic energy, which, by its definition, is true. However, if the car IS moving and there IS a force being applied to it, doesn't it have a change in kinetic energy?

Also, I got the homework back, and you're right about (a). The work done was 320N, but I'm just having a bit of trouble with the conceptual part of it.

5. Nov 7, 2012

### haruspex

Quite so. But if you were to apply the same logic you used originally to this case you would calculate no work was done.
That's useful work done, but the total work done includes losses to friction etc.

6. Nov 7, 2012

### Spinnor

You know the force and the mass so you can figure out the acceleration. You know the distance so you can figure out the time and the velocity.

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7. Nov 7, 2012

### PeterO

The implication of the "just keep the car moving" section [in red above] is that the car will move with basically zero velocity - a real conundrum. Having got the car the 100m, the amount of work done would have been 390 x 100 [force by distance].

Fortunately they apply a greater force - and thus do even more work than that.

The net force you correctly calculated will result in an acceleration of this 1400kg car - and over a distance of 100m, you could use the usual motion formulae to calculate the speed when the car reaches the gas station. [answer to part (c)] and thus calculate (b)

Alternately, the 390N force [doing 390 x 100 J of work] would get the car there with "zero" speed.
The extra work done by the 420N force will be the kinetic energy of the car (b) so you can then calculate (c).