Question

A 1.5 kg block is placed on a frictionless surface and attached to a spring with a spring constant of 100 N/m. If the block is displaced by 0.2 m from its equilibrium position, what is the potential energy stored in the spring?

• A. $1.0 \, \text{J}$
• B. $2.0 \, \text{J}$
• C. $0.5 \, \text{J}$
• D. $3.0 \, \text{J}$

Hint:

The potential energy in a spring depends on the square of the displacement from the equilibrium position. Always use the correct spring constant and displacement in your calculations.

Answer 1

The Correct Option is A

The potential energy stored in a spring is given by the formula: $$PE = \frac{1}{2} k x^2$$ Where: - $k = 100 \, \text{N/m}$ is the spring constant, - $x = 0.2 \, \text{m}$ is the displacement from the equilibrium position. Substituting the values: $$PE = \frac{1}{2} \times 100 \times (0.2)^2 = \frac{1}{2} \times 100 \times 0.04 = 1.0 \, \text{J}$$ Thus, the potential energy stored in the spring is $1.0 \, \text{J}$.