Question

question 11
problem reference 7.1
when jumping straight down, you can be seriously injured if you land stiff - legged. one way to avoid injury is to bend your knees upon landing to reduce the force of the impact. a 75.0 kg man just before contact with the ground has a speed of 6.40 m/s. in a stiff - legged landing he comes to a halt in 2.00 ms (milliseconds). when the man bends his knees, he comes to a halt in 0.100 s.
what is the average net force that acts on the man when he bands his kness?
4,800 n
3,750 n
375 n
48.0 n

Explanation:

Step1: Calculate the initial momentum

The formula for momentum is \(p = mv\). Given \(m = 75.0\space kg\) and \(v=6.40\space m/s\), the initial momentum \(p_{i}=mv=(75.0\space kg)\times(6.40\space m/s) = 480\space kg\cdot m/s\). The final momentum \(p_{f} = 0\space kg\cdot m/s\) (since he comes to a halt).

Step2: Use the impulse - momentum theorem

The impulse - momentum theorem is \(J=\Delta p=p_{f}-p_{i}\) and also \(J = F_{avg}\Delta t\). We know \(\Delta p=0 - 480\space kg\cdot m/s=- 480\space kg\cdot m/s\) and \(\Delta t = 0.120\space s\) (converting \(2.00\space ms=2\times10^{- 3}\space s\) is wrong, should be \(t = 0.120\space s\) if we use the correct formula, assume the problem has a typo in time unit, if \(t = 0.120\space s\)). Then \(F_{avg}=\frac{\Delta p}{\Delta t}\). Substituting \(\Delta p=-480\space kg\cdot m/s\) and \(\Delta t = 0.120\space s\), we get \(F_{avg}=\frac{- 480\space kg\cdot m/s}{0.120\space s}=-4000\space N\) (magnitude is \(4000\space N\)), if we use \(t = 0.100\space s\) (maybe a mis - write in the problem statement for the time when bending knees), \(F_{avg}=\frac{\Delta p}{\Delta t}=\frac{0 - 480}{0.100}=-4800\space N\) (magnitude \(F_{avg}=4800\space N\))

Answer:

4,800 N