A mass of 0.75 kg is suspended from a spring.If 1.41 J of work is done against the spring as the mass is pulled 0.15 m away from its equilibrium position,what is the spring constant of the spring?

Question

Quizplus · Accepted Answer

The potential energy stored in a spring is given by: PE = (1/2)kx^2, where k is the spring constant and x is the displacement from equilibrium position. 
In this case, the work done against the spring is equal to the change in potential energy: 
W = PE_final - PE_initial 
1.41 J = (1/2)k(0.15^2) - (1/2)k(0)^2 
1.41 J = (1/2)k(0.0225) 
k = (2*1.41 J)/(0.0225) 
k = 125 N/m, 
therefore the answer is (C) 125 N/m.

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