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# Edexcel M2 - Work done watch

1. A car of mass 1000 kg is towing a trailer of mass 1500 kg along a straight horizontal road. The two-bar joining the car to the trailer is modelled as a light rod parallel to the road. The total resistance to motion of the car is modelled has having constant magnitude 750 N. The total resistance to motion of the trailer is a force of 1250N.

When travelling at 25 m s-1 the driver of the car disengages the engine and applies the brakes. The brakes provide a constant braking force of magnitude 1500 N to the car. The resisting forces of magnitude 750 N and 1250 N are assumed to remain unchanged.

Calculate the work done, in kJ, by the braking force in bringing the car and the trailer to rest.

I understand that the solution is to use W=Fd, but why does using 1/2mv^2 give the incorrect answer?
2. (Original post by ambershell27)
A car of mass 1000 kg is towing a trailer of mass 1500 kg along a straight horizontal road. The two-bar joining the car to the trailer is modelled as a light rod parallel to the road. The total resistance to motion of the car is modelled has having constant magnitude 750 N. The total resistance to motion of the trailer is a force of 1250N.

When travelling at 25 m s-1 the driver of the car disengages the engine and applies the brakes. The brakes provide a constant braking force of magnitude 1500 N to the car. The resisting forces of magnitude 750 N and 1250 N are assumed to remain unchanged.

Calculate the work done, in kJ, by the braking force in bringing the car and the trailer to rest.

I understand that the solution is to use W=Fd, but why does using 1/2mv^2 give the incorrect answer?
The loss in KE will be equal to the work done against resistance. But there is more resistance than just the braking force.

3. (Original post by Notnek)
The loss in KE will be equal to the work done against resistance. But there is more resistance than just the braking force.

So if I understand correctly, 1/2mv^2 gives the work done by the whole system as opposed to the work done by the braking force?
4. (Original post by ambershell27)
So if I understand correctly, 1/2mv^2 gives the work done by the whole system as opposed to the work done by the braking force?
In this particular question yes.

1/2mv^2 is the kinetic energy. By the principle of work/energy, the total loss in mechanical energy (KE + PE) is equal to the total work done against resistance. In this case there is no PE so the total loss in KE will be equal to the work done against resistance.
5. Okay, that makes things clearer. Thanks

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