Question

Tarzan, whose mass is 94 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 2.8 m above the ground and the bottom of his dangling feet are at a height 2.0 above the ground. When he first hits the ground he has dropped a distance 2.0, so his center of mass is (2.8 - 2.0) above the ground. Then his knees bend and he ends up at rest in a crouched position with his center of mass a height 0.5 above the ground.(a) Consider the point particle system. What is the speed v at the instant just before Tarzan's feet touch the ground? v = _______ m/s. (b) Consider the extended system. What is the net change in internal energy for Tarzan from just before his feet touch the ground to when he is in the crouched position?

Answer 1

Answer:

(a) 6.26 m/s

(b) ‭-2,118.76 J

Explanation:

Here we have

Tarzan's mass, m = 94 kg

Height of feet above ground, h₁ = 2.0 m

Height of center of mass above ground = 2.8 m

Height of center of mass on the ground, h₂ = 2.8 - 2.0 = 0.8 m

Height of center of mass in the crouched position, h₃ = 0.5 m

(a) The speed at the instant just before Tarzan's feet touches the ground is given by;

v² = u² + 2·g·h₁

v = Speed at the instant just before Tarzan's feet touches the ground

u = Initial speed = 0 m/s while hanging from the tree

g = Acceleration due to gravity

Therefore, v² =  2·g·h₁ = 2 × 9.8 × 2 = 39.2 m²/s²

∴ v = √(39.2 m²/s²) = 6.26 m/s

(b) Here we have

Just before Tarzan's feet touches the ground internal energy is given by;

Initial Internal energy = K.E. + P.E. = m·g·h₂+ 0.5·m·v²

= 94 × 9.8 × 0.8 + 0.5 × 94 × 39.2 = ‭2,579.36 J

When in the crouched position, the final internal energy is given by;

m·g·h₃ = 94 × 9.8 × 0.5 = 460.6 J

Therefore net change in internal energy, ΔU is given by

ΔU = Final internal energy - Initial internal energy

ΔU = 460.6 J - 2,579.36 J  = ‭-2,118.76 J.