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
[tex]a=\frac{v-u}{t}[/tex]
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,
[tex]a=\frac{0-10 m/s}{0.26 s}=-38.5 m/s^2[/tex]
The force exerted on the passenger is given by the product between the mass of the passenger (70 kg) and the acceleration:
[tex]F=ma=(70 kg)(-38.5 m/s^2)=-2,695 N[/tex]
where the negative sign means that the direction of the force is opposite to the motion of the passenger+car.
We have that for the Question,it can be said that the force the seat belt exerts on a passenger in the car to bring him to a halt is
F_net=2.7*10^3
From the question we are told
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
Generally the equation for the net Force is mathematically given as
[tex]F_net=\frac{mdv}{dt}\\\\\\Therefore\\\\\F_net=\frac{70*10}{0.26}\\\\F_net=2.7*10^3\\\\\\[/tex]
Therefore
the force the seat belt exerts on a passenger in the car to bring him to a halt is
F_net=2.7*10^3
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