Question

A spring with spring constant 270 N/m has 5 J of energy stored in it. How much is it compressed?

Answer 1

Answer: 0.19 m

SOLUTION:

We can use the formula of potential energy stored in the spring, i.e

`\longrightarrow\underline{ \underline{ \sf { P.E. = (1)/(2) kx^2}}}`

where,

• (PE) represents potential energy.
• (k) represents spring constant
• (x) represents displacement or compression of the spring.

Also given that the spring constant (k) is 270 N/m, potential energy (PE) is 5 J.

To find (x), substitute the given values in the above formula,

`\longrightarrow \: \: \sf 5 = (1)/(2) * 270 * x^2 \\ \\ \longrightarrow \sf 5 = 135 * x^2 \\ \\ \longrightarrow \sf (5)/(135) = x^2 \\ \\ \longrightarrow \sf 0.0370 = x^2 \\ \\ \longrightarrow \sf x = \sqrt {0.0370} \\ \\ \longrightarrow {\underline{\underline{\sf {x \approx 0.19 m}}}}`

Hence, the spring is compressed 0.19 m approximately.