Impulse and momentum / work-energy

  • #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!
 
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  • #2
Welcome!

Consider that while the front of the car is being compressed by 0.94 m, its velocity is simultaneously and uniformly reduced from 70 km/h to zero.
Therefore, your calculated time is too short.
 
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  • #3
Oh ok that makes sense. Thanks!
 
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  • #4
Please use two \$$ to bracket LaTeX expressions. Click on "LaTeX Guide", lower left for more useful informaton about using LaTeX on this site.
The expression $$F = \frac{dp}{dt} \rightarrow F = m\frac{dv}{dt} \rightarrow F_{\text{avg}}\Delta t = m \Delta v$$ is appropriate for finding the average force if you know ##\Delta t## which you don't. In that case you have to make the additional assumption that the force is constant. The solution you have does that and calculates the work done by that force and averages over distance.

You could have gotten the same answer if you found ##\Delta t## by dividing ##\Delta x## not by the initial velocity ##v_0## but by the average velocity ##v_{\text{avg}}=\frac{1}{2}(v_0+0).## That's where your missing factor of 2 is. The answers are the same because the average velocity is calculated under the assumption that the acceleration (and hence the force) is constant.
 
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  • #5
One can dispense with the uniform acceleration requirement if one decides to creatively interpret the "average" force as an average over displacement rather than as an average over time. The two averages can differ in general. If the deceleration is constant, they will match.

We have a force that acts on the passenger over a known displacement while doing a known amount of work.

The average (over displacement) force is defined as $$\frac {\int f(s) \cdot ds}{\int ds} = \frac{\int f(s) \cdot ds}{\Delta s}$$ The work done is given by ##\int f(s) \cdot ds##. We know initial and final kinetic energies for the passenger, so we know the work done. That gives us the numerator of the fraction. We are given the total displacement, ##\Delta s##. So we know the denominator of the fraction.

With this method, the factor of ##\frac{1}{2}## comes from the formula for kinetic energy: ##KE=\frac{1}{2}mv^2##.
 
  • #6
jbriggs444 said:
One can dispense with the uniform acceleration requirement if one decides to creatively interpret the "average" force as an average over displacement rather than as an average over time.
While it is technically true that one can choose to define an average wrt one or another variable, it is highly misleading to intend average force as being wrt displacement without saying so. The default meaning of "average force" has to be average wrt time in order to be consistent with the default meaning of average acceleration. When did you last see a question asking for average acceleration wrt distance? Educators should know better.
That said, I cannot think what the practical value could be of computing an average force. In the real world, we care about momentum, energy and forces exceeding thresholds. So maybe no great harm done.
 

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