Question

A woman (mass= 50.5 kg) jumps off of the ground, and comes back down to the ground at a velocity of -8.4 m/s.

Rather than landing stiff-legged, it would be much safer for the woman to land with bent knees, which would bring her to a stop in about 0.27 seconds. Show the average net force that would act on her with this longer stopping time.

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Hint: You will need to find the change in the woman's momentum before and after she lands, and set that equal to our impulse formula ( p = F*t) to find the average force during the impact

Answer 1

Approximately
`1.6* 10^(3)\; \rm N`.

Step-by-step explanation:

By the Impulse-Momentum Theorem, the change in this woman's momentum will be equal to the impulse that is applied to her.

The momentum
`p` of an object is equal to the product of its mass
`m` and velocity
`v`. That is:
`p = m \cdot v`.

Let
`v(\text{before})` and
`v(\text{after})` represent the velocity of the woman before and after the landing. Let
`m` represent the woman's mass.

• The woman's momentum before the landing would be
  `m \cdot v(\text{before})`.
• The woman's momentum after the landing would be
  `m \cdot v(\text{after})`.

Therefore, the change in this woman's momentum would be:

`\begin{aligned}& \Delta p \\ & = p(\text{after}) - p(\text{before}) \\ &= m \cdot (v(\text{after})- v(\text{before}))\end{aligned}`.

On the other hand, impulse is equal to force multiplied by the duration of the force. Let
`F` represent the average force on the woman. The impulse on her during the landing would be
`F \cdot t`.

Apply the Impulse-Momentum Theorem.

• Impulse:
  `F\cdot t`.
• Change in momentum:
  `m \cdot (v(\text{after})- v(\text{before}))`.

Impulse is equal to the change in momentum:

`F \cdot t = m \cdot (v(\text{after})- v(\text{before}))`.

After landing, the woman comes to a stop. Her velocity would become zero. Therefore,
`v(\text{after}) = 0\; \rm m \cdot s^(-1)`.

`\begin{aligned}F &= \displaystyle \frac{m \cdot (v(\text{after})- v(\text{before}))}{t} \\ &= \frac{50.5\; \text{kg} * \left(0 \; \mathrm{m \cdot s^(-1)}- 8.4\; \mathrm{m \cdot s^(-1)}\right)}{0.27\; \rm s} \\ &\approx 1.6 * 10^(3)\; \rm N\end{aligned}`.