Question

A 60.0-kg man jumps 1.70 m down onto a concrete walkway. His downward motion stops in 0.025 seconds. If he forgets to bend his knees, what force (N) is transmitted to his leg bones?

Answer 1

Answer:

The magnitude of the force transmitted to his leg bones is 1413.6 N

Explanation:

Recall that force is defined as the change in linear momentum per unit time:

$F=\frac{P_f-P_i}{\Delta t}$

We can use this formula to find the force transmitted to his legs. We know that the final momentum ($P_f$) is 0 since the person is not moving on the floor, but we need to find what the person's momentum was an instant before he touches the ground. Since we know he person's mass, all we need for the initial momentum is his velocity.

For such we use conservation of energy in free fall, knowing that he jumped from 1.7 meters:

$Potential \,\,Energy \,\,at \,\,the \,\,top \,\,of \,\,the\,\, jump = U_i=m * g * h = 60*9.8*1.7 \,J\\Kinetic \,\,Energy \,\,when \,\,touching \,\,ground =KE_f= \frac{1}{2} m*v^2=\frac{60\,kg}{2} v^2\\\\KE_f=U_i\\\frac{60\,kg}{2} v^2=60*9.8*1.7 \,J\\v^2=2*9.8*1.7 \,\frac{m^2}{s^2} \\v=0.589\,\frac{m}{s}$

Now with this velocity, we know the $P_i$ (initial momentum) just before impact.

$P_i=60 \,kg * 0.589 \frac{m}{s} =35.34 \,kg\,\frac{m}{s}$

And since the impact lasted 0.025 seconds, we can find the force using the first formula we recalled:

$F=\frac{P_f-P_i}{\Delta t}=\frac{0-35.34_i}{\0.025}: N= -1413.6\,N$

so the magnitude of the force is 1413.6 N