Physics, 30.01.2020 04:46 kkelley3141

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-kg man just before contact with the ground has a speed of 6.4 m/s. (a) in a stiff-legged landing he comes to a halt in 2.0 ms. find the average net force that acts on him during this time. (b) when he bends his knees, he comes to a halt in 0.10 s. find 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 force of the ground on the man in parts (a) and (b).

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Answer from: kaylaamberd

(a) 240,000 N

We can solve the problem by using the impulse theorem, which states that the impulse exerted on the man is equal to its change in momentum:

$I=\Delta p\\F\Delta t= m(v-u)$

where

F is the average net force exerted on the man

$\Delta t=2.0 ms=0.002 s$ is the contact time

m = 75 kg is the man's mass

v=0 m/s is the final speed of the man

u = 6.4 m/s is his initial speed

Solving the formula for F, we find

$F=\frac{m(v-u)}{\Delta t}=\frac{(75 kg)(0-(6.4 m/s)}{0.002 s}=-240,000 N$

And the negative sign simply means that the force (upward) is opposite to the initial velocity of the man (downward).

(b) 4800 N

The problem can be solved exactly as the previous part, but this time the time of contact is

$\Delta t=0.10 s$

So the average force is

$F=\frac{m(v-u)}{\Delta t}=\frac{(75 kg)(0-(6.4 m/s)}{0.10 s}=-4800 N$

The force found in part (a) and (b) corresponds to the net force, which is the difference between the force exerted by the ground on the man (N, upward), and the weight of the man (W, downward):

F=N-W

For part (a), we can find N as

N=F+W=240,000 N-(mg)=240,000 N-(75 kg)(9.8 m/s^2)=239,265 N

While for part (b), we find

N=F+W=4800 N-(mg)=4800 N-(75 kg)(9.8 m/s^2)=4065 N

Answer from: Quest

Try c ? that’s what i got
An object is located 5.0 cm from a convex lens. the lens focuses light at a distance of 10.0 cm. wha

Answer from: Quest

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

$x(t)=ax(x-6)$

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at $x=0$ and ends at $x=0$ , so its displacement is 0.

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When jumping straight down, you can be seriously injured if you land stiff-legged. one way to avoid...