You hang by your hands from a tree limb that is a height L = 3.4 m above the gro

Question

You hang by your hands from a tree limb that is a height L = 3.4 m above the ground, with your center of mass a height h = 2.5 m above the ground and your feet a height d = 1.5 m above the ground, as shown in the figure (not to scale). You then let yourself fall. You absorb the shock by bending your knees, ending up momentarily at rest in a crouched position with your center of mass a height b = 0.3 m above the ground. Your mass is M = 56 kg. You will need to draw labeled physics diagrams for the various stages in the process. (a) What is the net internal energy change delta E_int in your body (chemical plus thermal)? (b) What is your speed v at the instant when your feet first touch the ground? V = m/s (c) What is the magnitude F of the approximate average force exerted by the ground on your feet during the time when your knees are bending? (The force of the Earth is only Mg = 549 N; you notice the unusually big force on your feet.) F = N (d) How much work is done by this force F? W = J

Explanation / Answer

a)

change in internal energy = m * g *(d -b)

change in internal energy = 56 * 9.8 *(1.5 - 0.3)

change in internal energy = 658.56 J

b)

let the speed is v

v = sqrt(2 * g * (1.5))

v = 5.42m/s

the speed is 5.42 m/s

c)

let the acceleration is a

a = v^2/(2 * d)

a = 5.42^2/(2 * (2.5 - 1.5 - 0.3))

a = 20.98 m/s^2

seocnd law of motion

N - Mg = M* a

N = 549 + 20.98 * 56

N = 1723.88 N

the magnitude of force F is 1723.88 N

d)

work done by this force = -F * (2.5 - 1.5 - 0.3)

work done by this force = -1723.88 * 0.7

work done by this force = - 1206.72 J

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