Question

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 60.0-kg man just before contact with the ground has a speed of 4.18 m/s. (a) In a stiff-legged landing he comes to a halt in 1.00 ms. Find the magnitude of the average net force that acts on him during this time. (b) When he bends his knees, he comes to a halt in 0.245 s. Find the magnitude of the average net force now. (c) During the landing, the force of the ground on the man points upward, while the force due to gravity points downward. The average net force acting on the man includes both of these forces. Taking into account the directions of the forces, find the magnitude of the force applied by the ground on the man in part (b).

Answer 1

a) The average force that acts on the man is
`2.508* 10^(8)` newtons.

b) The average force that acts on the man is 1023.673 newtons.

c) The force of the ground on the man is 1612.093 newtons upwards.

Step-by-step explanation:

a) After a careful reading of the statement we construct the following model by applying Impact Theorem, that is:

`m\cdot \vec v_(A) + \vec F \cdot \Delta t = m\cdot \vec v_(B)` (Eq. 1)

Where:

`m` - Mass of the man, measured in kilograms.

`\vec v_(A)` - Initial velocity of the man, measured in meters per second.

`\vec v_(B)` - Final velocity of the man, measured in meters per second.

`\Delta t` - Impact time, measured in seconds.

`\vec F` - Average net force, measured in newtons.

Now we proceed to clear average net force within expression:

`\vec F \cdot \Delta t = m\cdot (\vec v_(B)-\vec v_(A))`

`\vec F = (m)/(\Delta t)\cdot (\vec v_(B)-\vec v_(A))` (Eq. 2)

If we know that
`m = 60\,kg`,
`\vec v_(A) = -4.18\,\hat{j}\,\,\,\left[(m)/(s) \right]`,
`\vec v_(B) = 0\,\hat{j}\,\,\,\left[(m)/(s) \right]` and
`\Delta t = 1* 10^(-6)\,s`, we obtain the following vector:

`\vec F = (60\,kg)/(1* 10^(-6)\,s) \cdot (4.18\,\hat{j})\,\,\,\left[(m)/(s) \right]`

`\vec F = 2.508* 10^(8)\,\hat{j}\,\,\,[N]`

The average force that acts on the man is
`2.508* 10^(8)` newtons.

(b) If we know that
`m = 60\,kg`,
`\vec v_(A) = -4.18\,\hat{j}\,\,\,\left[(m)/(s) \right]`,
`\vec v_(B) = 0\,\hat{j}\,\,\,\left[(m)/(s) \right]` and
`\Delta t = 0.245\,s`, we obtain the following vector:

`\vec F = (60\,kg)/(0.245\,s) \cdot (4.18\,\hat{j})\,\,\,\left[(m)/(s) \right]`

`\vec F = 1023.673\,\hat{j}\,\,\,\left[N\right]`

The average force that acts on the man is 1023.673 newtons.

(c) From Second Newton's Law we find the following equation of equilibrium:

`\vec F = \vec N -\vec W` (Eq. 3)

Where:

`\vec F` - Average force that acts on the man, measured in newtons.

`\vec N` - Force of the ground on the man, measured in newtons.

`\vec W` - Weight of the man, measured in newtons.

By applying the concept of weight, we expand the previous equation:

`\vec F = \vec N -m\cdot \vec g` (Eq. 3b)

Where
`\vec g` is the gravitational acceleration, measured in meters per square second.

And then we clear the force of the ground on the man:

`\vec N = \vec F +m\cdot \vec g` (Eq. 4)

If we get that
`\vec F = 1023.673\,\hat{j}\,\,\,\left[N\right]`,
`m = 60\,kg` and
`\vec g = 9.807\,\hat{j}\,\,\,\left[(m)/(s^(2)) \right]`, the average force is:

`\vec N = 1023.673\,\hat{j}\,\,\,[N]+(60\,kg)\cdot (9.807\,\hat{j})\,\,\,\left[(m)/(s^(2)) \right]`

`\vec N = 1612.093\,\hat{j}\,\,\,\left[N\right]`

The force of the ground on the man is 1612.093 newtons upwards.