The jumping gait of the kangaroo is efficient because energy is stored in the st

Question

The jumping gait of the kangaroo is efficient because energy is stored in the stretch of stout tendons in the legs; the kangaroo literally bounces with each stride. We can model the bouncing of a kangaroo as the bouncing of a mass on a spring, A 70 kg kangaroo hits the ground, the tendons stretch to a maximum length, and the rebound causes the kangaroo to leave the ground approximately 0.1 s after its feet first touch. 8. KANGAROOS AREUST IHINKABOUTIT Modeling the gn-the-ground part (not the in-the-air part) of this motion as a mass on a spring, what is the effective period of the motion? (Hint: the point where the kangaroo touches/leaves the ground is the equilibrium point of the spring, since the tendon is unstretched there) 13 pt what is the effective spring constant of the tendon based on this model? [4 pt] If the kangaroo speeds up, it must bounce higher and farther with each stride, and so must store more energy in each bounce. Modeling it as an ideal spring, how would this affect the amplitude of tendon stretch? Explain by referring to an equation. [3 ptj If the kangaroo speeds up, it must bounce higher and farther with each stride, and so must store more energy in each bounce. Modeling it as an ideal spring, how would this affect the on-the-ground time for the bounce? Explain by referring to an equation. [2 pt] a. b, c. d.

Explanation / Answer

mass, m=70 kg

t=0.1 sec

a)

T=2*t

T=2*0.1

T=0.2 sec

b)

W=2pi/T

sqrt(K/m)=2pi/T

sqrt(k/70)=2pi/(0.2)

===> K=69087.23 N/m

c)

ampitude will increase

d)

time will remains same

T=2pi*sqrt(m/K), here T depends on only m and K

therefore,

time will remain same

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