Answer
We assume downward to be in positive direction.
a) The average net force acting on the man is $-2.4\times10^5N$
b) The average net force acting on the man is $-4.8\times10^3N$
c) In a), the force acted on the man by the ground is $-2.41\times10^5N$
In b), the force acted on the man by the ground is $-5.54\times10^3N$
Work Step by Step
We assume downward to be in positive direction.
According to the impulse-momentum theorem, $$\sum\vec{\overline F}\Delta t=m(\vec{v_f}-\vec{v_0})$$ $$\sum\vec{\overline F}=\frac{m(\vec{v_f}-\vec{v_0})}{\Delta t}$$
a) The man's mass $m=75kg$, $v_0=6.4m/s$, $v_f=0$ and $\Delta t=2ms=2\times10^{-3}s$
The average net force acting on the man is $$\sum\vec{\overline F}_a=-2.4\times10^5N$$
b) $m=75kg$, $v_0=6.4m/s$, $v_f=0$ and $\Delta t=0.1s$
The average net force acting on the man is $$\sum\vec{\overline F}_b=-4.8\times10^3N$$
c) The man's weight, which points downward, is $W=+(75\times9.8)=+735N$
Let's call the force exerted on the man by the ground $F$. We have $\sum\vec{\overline F}=\vec{F}+\vec{W}$
- In a), the force on the man due to the ground is $\vec{F}=\sum\vec{\overline F}_a-\vec{W}=-2.41\times10^5N$
- In b), the force on the man due to the ground is $\vec{F}=\sum\vec{\overline F}_b-\vec{W}=-5.54\times10^3N$
