Question

A kangaroo has a maximum gravitational potential energy during one jump of 770 J

Answer 1

Answer:

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;

$E.P.E = \frac {1}{2}kx^{2}$

Substituting into the equation, we have;

$107.8 = \frac {1}{2}*k*0.07^{2}$

$107.8 = \frac {1}{2}*k*0.0049$

Cross-multiplying, we have;

$215.6 = k*0.0049$

$k = \frac {215.6}{0.0049}$

k = 44000 N/m