College Physics by Openstax Chapter 8 Problem 8

A car moving at 10.0 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 70.0 kg.

Solution:

Newton’s second law of motion in terms of momentum states that the net external force equals the change in momentum of a system divided by the time over which it changes. In symbols, Newton’s second law of motion is defined to be

Fnet=ΔpΔt\text{F}_\text{net} = \frac{\Delta \textbf{p}}{\Delta t}

Moreover, Linear momentum (momentum for brevity) is defined as the product of a system’s mass multiplied by its velocity.

p=mv\textbf{p} = m \textbf{v}

So, the Newton’s second law becomes

Fnet=mΔvΔt\text{F}_\text{net} = \frac{m\Delta\textbf{v}}{\Delta t}

Substituting the given values, we have

Fnet=mΔvΔtFnet=(70 kg)(10 m/s)0.26 sFnet=2692.3077 NFnet=2.69×103 N  (Answer)\begin{align*} \text{F}_\text{net} = & \frac{m \Delta\textbf{v}}{\Delta t} \\ \text{F}_\text{net} = & \frac{\left( 70\ \text{kg} \right)\left( 10\ \text{m/s} \right)}{0.26\ \text{s}} \\ \text{F}_\text{net} = & 2692.3077 \ \text{N} \\ \text{F}_\text{net} = & 2.69 \times 10^{3}\ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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