Solve the problem.
-A spring on a horizontal surface can be stretched and held 0.5 m from its equilibrium position with a force of 45 N. How much work is done in stretching the spring 3 m from its equilibrium position?
A) 67.5 J
B) 405 J
C) 135 J
D) 810 J

Question

Solve the problem.   
-A spring on a horizontal surface can be stretched and held 0.5 m from its equilibrium position with a force of 45 N. How much work is done in stretching the spring 3 m from its equilibrium position?
A) 67.5 J
B) 405 J
C) 135 J
D) 810 J

Quizplus · Accepted Answer

The work done on a spring is given by the formula $W = \frac{1}{2}kx^2$, where $k$ is the spring constant and $x$ is the displacement from the equilibrium position. First, we find $k$ using the given force and displacement: $F = kx$, so $k = \frac{F}{x} = \frac{45\,N}{0.5\,m} = 90\,N/m$. Then, we calculate the work for a 3 m displacement: $W = \frac{1}{2}(90\,N/m)(3\,m)^2 = \frac{1}{2}(90)(9) = 405\,J$.

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