Question

A truck moving at 13.3 m/s hits a concrete wall. As a result of the collision, a 6-kg wrench moves forwards and strikes the wall of the tool compartment. If the wrench stops after being in contact with the wall for 0.07 s, what is the average force exerted on the wrench by the wall

Answer 1

Answer:

Explanation:

The velocity of the wrench must be equal to the velocity of the truck . So momentum of the wrench before it hits the wall

= mv = 6 x 13.3 = 79.8 kg m /s

If resisting force of wall be F , impulse on the wrench = F x time

= F x .07

Impulse = change in momentum of the wrench = mv - 0 = mv = 79.8 kgm/s

So F x .07 = 79.8

F = 1140 N .