Tarzan, whose mass is 103 kg, is hanging at rest from a tree limb. Then he lets go and falls to the ground. Just before he lets go, his center of mass is at a height 2.4 m above the ground and the bottom of his dangling feet are at a height 1.5 above the ground. When he first hits the ground he has dropped a distance 1.5, so his center of mass is (2.4 - 1.5) above the ground. Then his knees bend and he ends up at rest in a crouched position with his center of mass a height 0.3 above the ground.

Required:
Consider the real system. What is the net change in internal energy for Tarzan from just before his feet touch to the ground to when he is in the crouched position?

Question

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moya1316 moya1316 · Answer 1 · 2021-04-04T03:59:39+02:00

Answer:

v₀ = 60.38 mi / h

With this stopping distance, the starting speed should have been 60.38 mi/h, which is much higher than the maximum speed allowed.

Explanation:

For this exercise let's start by using Newton's second law

Y axis

N-W = 0

N = W

X axis

fr = m a

the expression for the friction force is

fr = μ N

we substitute

μ mg = m a

μ g = a

calculate us

a = 0.620 9.8

a = 6.076 m / s²

now we can use the kinematics relations

v² = v₀² - 2 a x

suppose v = 0

v₀ = [tex]\sqrt{2ax}[/tex]Ra 2ax

let's calculate

v₀ = [tex]\sqrt{2 \ 6076 \ 60}[/tex]

v₀ = 27.00 m / s

let's slow down to the english system

v₀ = 27.0 m / s (3.28 ft / 1m) (1 mile / 5280 ft) (3600s / 1h)

v₀ = 60.38 mi / h

With this stopping distance, the starting speed should have been 60.38 mi/h, which is much higher than the maximum speed allowed.