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a car moving at 12 m/s crashes into a tree and stops in 0.24 s. calculate the magnitude of the average force in newtons which the seat belt exerts on a passenger in the car to bring him to a halt. the mass of the passenger is 60 kg.

Respuesta :

The force which the seat belt exerts on a passenger is - 3000 N. Here minus sign indicates force was against the motion.

Given ;  The velocity is v = 12 m/s

and the time = 0.24 sec

Since he crashes in the end , thus the final velocity = 0 m/s

Weight = 60 kg

To calculate : The average force

We will use the formula :

ΔP = FΔT

p = mv

that is initial momentum = final momentum

F = [tex]\frac{m(final velcoity - initial velocity)}{time}[/tex]

F = 60(0 - 12 ) / 0.24

On further solving we get

F = - 720 / 0.24

F = - 3000N

Thus the force which the seat belt exerts on a passenger is - 3000 N

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https://brainly.com/question/14697562

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