An 80kg red kangaroo jumps straight up in order to avoid a collision with an app

Question

An 80kg red kangaroo jumps straight up in order to avoid a collision with an approaching jeep. The 2m tall marsupial can crouch down until his center of mass is 0.5m above the ground. When his feet leave the ground (he can no longer push), his center of mass is 1.5m off the ground. He can push against the ground with a force 2.5 times its own weight. (He can really push much harder when he is leaping at an angle other than straight up, but I wanted to keep this one-dimensional, so this is an approximation!)

Open-Response Homework Problem 16.2 An 80kg red kangaroo jumps straight up in order to avoid a colli- marsupial can crouch down until his center of mass is 0.5m above the ground. When his feet leave the ground (he can no longer push), his center of mass is 1.5m off the ground. He can push against the ground with a force 2.5 times its own weight. (He can really push much harder when he is leaping at an angle other than straight up, but I l, so this is an appraximation!) sion with an approaching jeep. The 2m tall wanted to keep this (a)[5 pts If you were to identify just the kangaroo as your system, find the value of the net external force on the kangaroo as it is pushing. Make sure to draw a free body diagram on the figure to the right for the kangaroo during this time (while his feet (d) are in contact with the ground) the force pairs of the forces drawn on the kangaroo? pushing, what is its kinetic energy w (b)[2 pts What objects would you need to draw to include (c)[4 pts Assuming the kangaroo was at rest when it started hen its feet leave the (d)[7 pts] How much work did the ground do on the kanga- Carefully draw a work-energy diagram on the figure to ground? roo (write this down!)? the right showing how energy was transferred during the push for the defined system (the kangaroo). Work (e)[4 pts] How high is the kangaroo's center of mass from the ground at the top of the jump? (You may want to change systems here.])

Explanation / Answer

a) The Earth is outside the system, so it exerts an external force.The kangaroo’s weight is the force that the Earth’s gravity exertson it, or Fge,roo=mg= 784 N downward. The roo can push at 2.5 times this force, or Fcroo,ground= 1960 N, also downward (so Newton’s III Law says the ground pushes back on the roo withthis force,Fcground,roo, upward). The net external force on theroo (taking upward to be positive) is thus the force of the ground minus the force due to gravity,

Fext= (2.5 - 1)mg= 1176 N

Fge,roo =======> shown

Fcground,roo > Fge,roo

|Fcground,roo| = |2

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