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 65.3-kg man just before contact with the ground has a speed of 7.05 m/s. (a) In a stiff-legged landing he comes to a halt in 1.13 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 0.215 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)
I figured out a and b -
a) (65.3)(7.05) = (1.13x10^-3)(F)
b) (65.3)(7.05) = (.215)(F)
c) ???? =[
ground force-downwardforce=mass*acceleration
acceleration=changevelocity/time
So the downward force is mg s0 (63.5)(9.81) and is my ground force the F value from b? I subtract those two and then solve for acceleration and then F? I'm not sure which are the downward and ground forces..?
Oh wait. Is it (v)x(t) = a and then (a)x(m) = F?
the downward force is what you did in b, supposedly. You are looking for gound force. You can figure the acceleration in b.
acceleration= (velocity final-velocityinitla)/timeslowing
Are you overthinking?
So F = 2141 in b.
2141 = (65.3)(a)
a = 32.787
But part c asks for force, if i figure acceleration from part b, what does that do for me? Can you work with me with the numbers please?
force downward- force upward=mass*change acceleration.
force downward you figured in part b due to change in motion.
then solve for force upward to do the slowing in part b. now add his weight, which is also supported by the ground.