A car moving at 10 m/s crashes into a large bush and stops in 1.3 m. Using the Work-Energy theorem, calculate the average force the seat belt exerts on a passenger in the car to bring him to a halt. The mass of the passenger is 70 kg. How long does this take?

The magnitude of the average force the seat belt exerts on the passenger is 2692.3 N.

It takes 0.26 s to bring the passenger in the car to a halt.

Explanation:

Hi there!

The negative work (W) needed to bring a moving object to stop is equal to its kinetic energy (KE):

W = KE

F · s = 1/2 · m · v²

Where:

F = applied force on the passenger.

s = displacement of the passenger.

m = mass of the passenger.

v = velocity of the passenger.

Solving the equation for F:

F = 1/2 · m · v² / s

Replacing with the dа ta:

F = 1/2 · 70 kg · (10 m/s) ² / 1.3 m

F = 2692.3 N

The magnitude of the average force the seat belt exerts on the passenger is 2692.3 N.

According to the second law of Newton:

F = m · a

Where "a" is the acceleration of the passenger.

We also know from kinematics that the velocity of an object can ve calculated as follows:

v = v0 + a · t

Where:

v = velocity of the passenger at time t.

v0 = initial velocity.

t = time.

a = acceleration.

When the passenger stops, its velocity is zero. So replacing a = F/m, let's solve the equation for the time it takes the passenger to stop:

v = v0 + a · t

0 = 10 m/s + (-2692.3 N / 70 kg) · t

-10 m/s / (-2692.3 N / 70 kg) = t

t = 0.26 s

It takes 0.26 s to bring the passenger in the car to a halt.