A 2400 kg car moving at 11 m/s crashes into a tree and stops in 0.26 s. calculate the force the seat belt exerts on passenger in the car to bring him to a halt the mass of the passenger is 72 kg.
To calculate the force exerted by the seat belt on the passenger, we can use Newton's second law of motion, which states that F = ma, where F is the force, m is the mass, and a is the acceleration.
First, let's find the acceleration of the car using the formula for acceleration:
a = (final velocity - initial velocity) / time
Given:
Initial velocity (u) = 11 m/s
Final velocity (v) = 0 m/s
Time (t) = 0.26 s
Using the formula, we have:
a = (0 - 11) m/s / 0.26 s
a = -11 m/s / 0.26 s
a = -42.3077 m/s²
The negative sign indicates that the car experiences deceleration.
Now, we can calculate the force exerted on the car to bring it to a halt. Since there are two forces acting on the car (the force exerted by the seat belt and the force of friction with the tree), they add up to zero as the car comes to a stop. Therefore, the force exerted by the seat belt is equal in magnitude and opposite in direction to the force of friction.
To calculate the force exerted by the seat belt, we need to determine the force of friction acting on the car. We can do this using Newton's second law:
Force of friction = mass × acceleration
Using the mass of the car and the calculated acceleration:
Force of friction = 2400 kg × (-42.3077 m/s²)
Force of friction = -101538.46 N
The negative sign indicates that the force of friction acts in the opposite direction to the motion of the car.
Since the force exerted by the seat belt is equal in magnitude and opposite in direction to the force of friction, the force exerted by the seat belt on the passenger is:
Force exerted by seat belt = -101538.46 N
Therefore, the seat belt exerts a force of -101,538.46 N on the passenger to bring him to a halt.
Note: The negative sign indicates that the force is directed opposite to the motion of the car.