Question

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.

Answer 1

Given,
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 = $\frac{1}{2}m(v^{2}-u^{2})$
=> Work Done = [text]\frac{1}{2}(3000(0^{2}-20^{2}))[/text]
=> Work Done = 600,000 J

Therefore, Work Done = 6 * $10^{5}$ J
We know, Work done = Fs (s =displacement)
=> 6 * $10^{5}$  = 24000 * s
=> s = 25m

Answer 2

Answer:

therefore s=25

Explanation:

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