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What is the declaration of a 2,200kg truck if a force of 4,000N is used to stop it?​

User PazO
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1 Answer

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The question does not give any frictional forces like air resistance so we have to assume that these are negligible.

The deceleration, a, of the car can be found using:

v v = u u + 2 a s

where the u is the initial speed of 8.9 m/s and v = o as the car comes to a stop.

The distance, s, it slows down over is 12 m.

So, 0 = u x u + 2 x a x 12

A bit of maths gives:

a = - 3.3 m/ss (the negative sign just tells us that the car is decelerating).

If friction is zero, then the braking force, F, is given by F = ma as this is the only horizontal force acting on the car, which we assume is on a horizontal road:

F = m a

So, F = 800 x - 3.3 = - 2640 N.

(Again the negative sign just tells us that the car is decelerating).

Note that your spelling of braking force in the question was wrong.
User Caren
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