Answer :
Answer:
a) The average force that acts on the man is [tex]2.508\times 10^{8}[/tex] 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.
Explanation:
a) After a careful reading of the statement we construct the following model by applying Impact Theorem, that is:
[tex]m\cdot \vec v_{A} + \vec F \cdot \Delta t = m\cdot \vec v_{B}[/tex] (Eq. 1)
Where:
[tex]m[/tex] - Mass of the man, measured in kilograms.
[tex]\vec v_{A}[/tex] - Initial velocity of the man, measured in meters per second.
[tex]\vec v_{B}[/tex] - Final velocity of the man, measured in meters per second.
[tex]\Delta t[/tex] - Impact time, measured in seconds.
[tex]\vec F[/tex] - Average net force, measured in newtons.
Now we proceed to clear average net force within expression:
[tex]\vec F \cdot \Delta t = m\cdot (\vec v_{B}-\vec v_{A})[/tex]
[tex]\vec F = \frac{m}{\Delta t}\cdot (\vec v_{B}-\vec v_{A})[/tex] (Eq. 2)
If we know that [tex]m = 60\,kg[/tex], [tex]\vec v_{A} = -4.18\,\hat{j}\,\,\,\left[\frac{m}{s} \right][/tex], [tex]\vec v_{B} = 0\,\hat{j}\,\,\,\left[\frac{m}{s} \right][/tex] and [tex]\Delta t = 1\times 10^{-6}\,s[/tex], we obtain the following vector:
[tex]\vec F = \frac{60\,kg}{1\times 10^{-6}\,s} \cdot (4.18\,\hat{j})\,\,\,\left[\frac{m}{s} \right][/tex]
[tex]\vec F = 2.508\times 10^{8}\,\hat{j}\,\,\,[N][/tex]
The average force that acts on the man is [tex]2.508\times 10^{8}[/tex] newtons.
(b) If we know that [tex]m = 60\,kg[/tex], [tex]\vec v_{A} = -4.18\,\hat{j}\,\,\,\left[\frac{m}{s} \right][/tex], [tex]\vec v_{B} = 0\,\hat{j}\,\,\,\left[\frac{m}{s} \right][/tex] and [tex]\Delta t = 0.245\,s[/tex], we obtain the following vector:
[tex]\vec F = \frac{60\,kg}{0.245\,s} \cdot (4.18\,\hat{j})\,\,\,\left[\frac{m}{s} \right][/tex]
[tex]\vec F = 1023.673\,\hat{j}\,\,\,\left[N\right][/tex]
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:
[tex]\vec F = \vec N -\vec W[/tex] (Eq. 3)
Where:
[tex]\vec F[/tex] - Average force that acts on the man, measured in newtons.
[tex]\vec N[/tex] - Force of the ground on the man, measured in newtons.
[tex]\vec W[/tex] - Weight of the man, measured in newtons.
By applying the concept of weight, we expand the previous equation:
[tex]\vec F = \vec N -m\cdot \vec g[/tex] (Eq. 3b)
Where [tex]\vec g[/tex] is the gravitational acceleration, measured in meters per square second.
And then we clear the force of the ground on the man:
[tex]\vec N = \vec F +m\cdot \vec g[/tex] (Eq. 4)
If we get that [tex]\vec F = 1023.673\,\hat{j}\,\,\,\left[N\right][/tex], [tex]m = 60\,kg[/tex] and [tex]\vec g = 9.807\,\hat{j}\,\,\,\left[\frac{m}{s^{2}} \right][/tex], the average force is:
[tex]\vec N = 1023.673\,\hat{j}\,\,\,[N]+(60\,kg)\cdot (9.807\,\hat{j})\,\,\,\left[\frac{m}{s^{2}} \right][/tex]
[tex]\vec N = 1612.093\,\hat{j}\,\,\,\left[N\right][/tex]
The force of the ground on the man is 1612.093 newtons upwards.