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A 0.5 kg brick is dropped from a height of 6.0 m. It hits the ground and comes to rest. (a) What is the magnitude of the impulse exerted by the ground on the brick? N·s (b) If it takes 0.0013 s from the time the brick first touches the ground until it comes to rest, what is the magnitude of the average force exerted by the ground on the brick?

User Zurgl
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1 Answer

3 votes

Answer:

a). Impulse = 32.373 Ns

b). Force = 2490 N

Step-by-step explanation:

Impulse = force * time

Force = Mgh

But time is unknown.

From

V² = U² + 2as

But V = at

(at)²= 0² + 2as

But a = 9.81. ( Acceleration of free fall)

t² =( 2*9.81*6)/96.2361

t = 1.1 sec

Force = Mgh

Force = 0.5*6*9.81

Force = 29.43 N

Impulse = 29.43*1.1

Impulse = 32.373 Ns

But if it takes 0.013 sec ...

Force = 32.373/0.013

Force = 2490 N

User Andreas Richter
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