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

lol

explanation:

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards