## A kangaroo has a maximum gravitational potential energy during one jump of 770 J Helppppp When the kangaroo lands on the ground

Question

A kangaroo has a maximum gravitational potential energy during one jump of 770 J
Helppppp
When the kangaroo lands on the ground 14% of the maximum gravitational potential energy is transferred to elastic potential energy in one tendon.The tendon has an unstretched length of 35.0 cm
When the kangaroo lands on the ground the tendon stretches to a length of 42.0 cm
Calculate the spring constant of the tendon.
[5 marks]
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3 years 2021-08-21T10:53:22+00:00 1 Answers 163 views 0

k = 44000 N/m

Explanation:

Given the following data;

Maximum gravitational potential energy = 770 J

Elastic potential energy = 14% of 770 J = 14/100 * 770 = 107.8 J

Extension, x = 42 – 35 = 7cm to meters = 7/100 = 0.07 m

To find the spring constant, k;

The elastic potential energy of an object is given by the formula;

Substituting into the equation, we have;

Cross-multiplying, we have;

k = 44000 N/m