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A rubber band stores 1.59 J of PEewhen it is stretched 0.292 m.What is its spring constant?(Unit=N/m)

User Cjmconie
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1 Answer

19 votes
19 votes

Given:

The potential energy stored in the spring is P.E. = 1.59 J

The spring is stretched by x = 0.292 m

To find the spring constant.

Step-by-step explanation:

The spring constant can be calculated by the formula


\begin{gathered} P.E.\text{ = }(1)/(2)kx^2 \\ k=(2* P.E.)/(x^2) \end{gathered}

On substituting the values, the spring constant will be


\begin{gathered} k\text{ =}\frac{2*1.59\text{ J}}{(0.292)^2m^2} \\ =37.3\text{ N/m} \end{gathered}

Thus, the spring constant is 37.3 N/m

User Tom Cabanski
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