A spring has a spring constant of 53 N/m. How much elastic potential energy is stored in the spring when it is compressed by 0.21 m?A.11.1 JB.1.17 JC.2.33 JD.5.57 JSUBMITarrow_backPREVIOUS
Question
A spring has a spring constant of 53 N/m. How much elastic potential energy is stored in the spring when it is compressed by 0.21 m?A.11.1 JB.1.17 JC.2.33 JD.5.57 JSUBMITarrow_backPREVIOUS
Solution
The elastic potential energy stored in a spring is given by the formula:
PE = 1/2 * k * x^2
where: PE is the potential energy, k is the spring constant, and x is the distance the spring is stretched or compressed.
Given: k = 53 N/m, x = 0.21 m.
Substituting these values into the formula, we get:
PE = 1/2 * 53 N/m * (0.21 m)^2 PE = 1/2 * 53 N/m * 0.0441 m^2 PE = 1.17 J
So, the correct answer is B.1.17 J.
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