Question

A spring with a spring constant of 27 N/m is stretched 16 cm. What is the spring's potential energy?

Answer 1

Given,

A spring with a spring constant of 27 N/m

To find,

Springs potential energy when stretched 16cm

Solution,

Potential energy in the spring due to a stretch $= \frac{1}{2}kx^{2}$

Where,  $k$ is the spring constant and $x$ is the change in length of the spring.

Here, $x = 16cm = 0.16m$

Spring constant $(k) = 27N/m$

Spring's potential energy  $= \frac{1}{2}kx^{2}$

                                            $= \frac{1}{2}*27*(0.16)^{2}\\= \frac{1}{2}*27*0.0256\\= 0.3456J$

Therefore, the energy stored is $0.3456J$

Answer 2

Answer:

Correct option is A)

Initial PE =0.5×k×x

2

=0.5×10×0.2

2

=0.2J

Final PE =0.5×k×x

2

=0.5×10×0.25

2

=0.3125J

Increase in PE =0.3125−0.2=0.1125≈0.1J

Hence correct answer is option A