Question

The force required to stretch a Hooke’s-law spring varies from 0 N to 63.5 N as we stretch the spring by moving one end 5.31 cm from its unstressed position. Find the force constant of the spring. Answer in units of N/m. Find the work done in stretching the spring. Answer in units of J.

Answer 1

Force constant will be 1195.85 N/m

Work done will be 1.6859 J

Step-by-step explanation:

We have given the force, F = 63.5 N

Spring is stretched by 5.31 cm

So x = 0.0531 m

Force is given , F = 63.5 N

We know that force is given by
`F=kx`

So
`63.5=k* 0.0531`

k = 1195.85 N/m

Now we have to find the work done

We know that work done is given by

`W=(1)/(2)kx^2=(1)/(2)* 1195.85* 0.0531^2=1.6859J`