Answer:
Explanation:
Given
Free fall acceleration on mars 
Time Period of pendulum on earth 
Time period of Pendulum is given by

for earth



(b)For same time period on mars length is given by



Newtons first law of motion is also known as the law of inertia
The potential energy of the spring is 6.75 J
The elastic potential energy stored in the spring is given by the equation:

where;
k is the spring constant
x is the compression/stretching of the string
In this problem, we have the spring as follows:
k = 150 N/m is the spring constant
x = 0.3 m is the compression
Substituting in the equation, we get


Therefore. the elastic potential energy stored in the spring is 6.75J .
Learn more about potential energy here:
brainly.com/question/10770261
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