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.

Force constant will be 1195.85 N/m

Work done will be 1.6859 J

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

So

k = 1195.85 N/m

Now we have to find the work done

We know that work done is given by

b) phosphorus pentoxide