Question

a kangaroo has a maximum gravitational potential energy of 770 joules. when the kangaroo lands on the ground 14% of the maximum gravitational potential energy is transferred to elastic potential energy in one tendon. A tendon has an unstretched length of 35 cm . When the kangaroo lands it stretches to 42 cm. calculate the spring constant of the tendon in meters.

Answer 1

$$0.5kx^2=\eta E_p\\0.5(0.42-0.35)^2k=0.14(770)\\k=44000\frac{N}{m}$$