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]
Show ur working out

Answers

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;

[tex] E.P.E = \frac {1}{2}kx^{2}[/tex]

Substituting into the equation, we have;

[tex] 107.8 = \frac {1}{2}*k*0.07^{2}[/tex]

[tex] 107.8 = \frac {1}{2}*k*0.0049 [/tex]

Cross-multiplying, we have;

[tex] 215.6 = k*0.0049 [/tex]

[tex] k = \frac {215.6}{0.0049} [/tex]

k = 44000 N/m

Answer 2

The spring constant of the tendon is 44000 N/m.

The following can be depicted from the question:

Elastic potential energy will be:

= 14% × 770J = 107.80 Joule.

The spring constant will be calculated thus:

107.8 = 1/2 × k × 0.07²

107.8 = 0.5 × k × 0.0049

k = 215.6 / 0.0049

k = 44000 N/m.

Therefore, the spring constant of the tendon is 44000 N/m.

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