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Mice21 [21]
3 years ago
11

A 1500 kg car moving with a speed of 20 m/s collides with a utility pole and is brought to rest in 0.30 s. Find the magnitude of

the average force exerted on the car during the collision.
Physics
1 answer:
storchak [24]3 years ago
7 0

Answer:

-100000 N.

Explanation:

Force: This can be defined as the product of the mass of a body and it's acceleration. The S.I unit of Force is Newton(N). The Formula of force is given as,

F = ma ........................... Equation 1

Where F = Average force exerted on the car, m = mass of the car, acceleration of the car, a = acceleration of the car.

a = (v-u)/t..................... Equation 2

Where v = Final velocity, u = Initial velocity, t = time.

Substitute equation 2 into equation 1

F = m(v-u)/t............. Equation 3

Given: m = 1500 kg, u = 20 m/s, v = 0 m/s (brought to rest), t = 0.3 s.

Substitute into equation 3

F = 1500(0-20)/0.3

F = 1500(-20)/0.3

F = -100000 N.

Note: The negative sign is due to the fact that the force exerted on the car by the pole is equal and opposite the force of the car.

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