- #1

GoonP

- 2

- 1

- Homework Statement
- A car moving at 70km/h collides rams into an immobile steel wall. Its front of the is

compressed by 0.94m . What average force must a seat belt exert in order to restrain

a 75-kg passenger?

- Relevant Equations
- F = \frac{dp}{dt}

Hello, I am wondering why this take at a solution is wrong.

So basically, $F = \frac{dp}{dt} \Rightarrow F = m\frac{dv}{dt} \Rightarrow F_{\text{avg}}\Delta t = m \Delta v$. Quick conversions show that 70 km/hr = 19.4 m/s, and therefore $\Delta t = \frac{0.94}{19.4} = 0.04845$. Therefore, the answer I am getting is $\frac{75 \cdot 19.4}{0.04845} \approx 30 \text{kN}$ which is exactly double the correct answer.

The solution I have access too uses the fact that $F_{\text{avg}} = \frac{\Delta E}{\Delta x}$ which i can understand is a basic consequence of work energy theorem, and substituting kinetic energy for $E$ it is easy to see mathematically why it is half my answer.

Any guidance as to why my initial approach is incorrect?

Thanks!

So basically, $F = \frac{dp}{dt} \Rightarrow F = m\frac{dv}{dt} \Rightarrow F_{\text{avg}}\Delta t = m \Delta v$. Quick conversions show that 70 km/hr = 19.4 m/s, and therefore $\Delta t = \frac{0.94}{19.4} = 0.04845$. Therefore, the answer I am getting is $\frac{75 \cdot 19.4}{0.04845} \approx 30 \text{kN}$ which is exactly double the correct answer.

The solution I have access too uses the fact that $F_{\text{avg}} = \frac{\Delta E}{\Delta x}$ which i can understand is a basic consequence of work energy theorem, and substituting kinetic energy for $E$ it is easy to see mathematically why it is half my answer.

Any guidance as to why my initial approach is incorrect?

Thanks!