# A 3000kg truck moving at a speed of 72km/hr stops after covering some distance. the force applied by the brakes is 24000N. compute the work and distance covered by this force.

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mass, m = 3000 kg

initial velocity, u = 72km/h = 20m/s

final velocity, v = 0m/s (Since it stops)

Force, F = 24000 N

Work Done = change in kinetic velocity =

=> Work Done = [text]\frac{1}{2}(3000(0^{2}-20^{2}))[/text]

=> Work Done = 600,000 J

Therefore, Work Done = 6 * J

We know, Work done = Fs (s =displacement)

=> 6 * = 24000 * s

=> s = 25m