Question

A load of 36 N attached to a spring hanging

Answer 1

Answer:

The required force is 55.38 N.

Explanation:

First we have to find the spring constant K;

                         k = $\frac{mg}{d}$ = $\frac{36}{0.065}$ = 553.846 N/m

As,

                         F= -K$x$

                   or, F=  - 553.846 × 0.10

                   or, F=  -55.3846 N

Hence,

                  $F_{needed}$ = - $F_{spring}$

            ∴  $F_{needed}$ = 55.38 N

The required force  to stretch it by this  amount is 55.38 N.