Question

A car moving at 10 m/s crashes into a tree and stops in 0.26 s. calculate the 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

Answer 1

Answer: -2695 N

Explanation:

first of all, we need to calculate the acceleration experienced by the car and the passenger. This acceleration is given by

$a=\frac{v-u}{t}$

where

v=0 is the final velocity of the car, after the impact

u=10 m/s is the initial velocity

t=0.26 s is the duration of the collision

Substituting numbers into the formula,

$a=\frac{0-10 m/s}{0.26 s}=-38.5 m/s^2$

The force exerted on the passenger is given by the product between the mass of the passenger (70 kg) and the acceleration:

$F=ma=(70 kg)(-38.5 m/s^2)=-2,695 N$

where the negative sign means that the direction of the force is opposite to the motion of the passenger+car.

Answer 2

A=(Vo-V)/t=(10-0)/0.26=10/0.26=38.46m/s2
F=ma=70*38.46=2692.3N