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
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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