Question

A force of 20N stretches a spring by 0.5m. The spring obeys Hooke’s law. This is a continuation of the previous question. b. How much force must a man use to stretch it by 1.5m?

Answer 1

Given:-

Force, F=20 N

deflection in a spring due to the force 20 N=0.5 m

final deflection, $\delta$=1.5 m

To Find:-

Force required to stretch spring by 1.5 m

Explanation:-

$\text{From the given data, Spring constant(k) of the spring}=\dfrac{20}{0.5}=40N/m\\\Rightarrow \text{Force exerted in the spring to stretch it by 1.5m}, F'=k\times\delta\\\Rightarrow F'=40\times1.5=60N$