Question

A spring gains 2.34 joules of elastic potential

energy as it is compressed 0.250 meter from its
equilibrium position. What is the spring constant
of this spring?
(1) 9.36 N/m (3) 37.4 N/m
(2) 18.7 N/m (4) 74.9 N/m

Answer 1

Elastic energy equals 1/2 k(spring constant) *Δx^2(change in position) so you set the energy of the spring equal to the 1/2kΔx^2 and solve for K. 2.34=1/2k(.250)^2 which gives you a spring constant of 74.9 N/m.

Answer 2

The correct answer to the question is: 4) 74.9 N/m.

Step-by-step explanation:

As per the question, the stretched length of the spring is given as x = 0.250 m.

The potential energy gained by the spring is given as 2.34 joules.

We are asked to calculate the spring constant of the spring.

The potential energy gained by the spring is nothing else than the elastic potential energy .

The elastic potential energy of the spring is calculated as -

Potential energy P.E =
`(1)/(2)kx^2`

⇒k =
`(2P.E)/(x^2)`

=
`(2* 2.34)/((0.250)^2)\ N/m`

= 74.88 N/m

= 74.9 N/m. [ans]

Hence, the force constant of the spring is 74.9 N/m.