Answer to Question #186915 in Mechanics | Relativity for TAYLOR

Question

A load of 37 N attached to a spring hanging vertically stretches the spring 3.4 cm. The spring is now placed horizontally on a table and stretched 15 cm. What force is required to stretch it by this amount? Answer in units of N.

Answer 1

Solution.

$F_1=37N;$

$x_1=3.4cm=0.034m;$

$x_2=15cm=0.15m;$

$F_2-?;$

$F=kx\implies k=\dfrac{F}{x};$

$k=\dfrac{F_1}{x_1};k=\dfrac{F_2}{x_2};$

$\dfrac{F_1}{x_1}=\dfrac{F_2}{x_2}\implies F_2=\dfrac{F_1x_2}{x_1};$

$F_2=\dfrac{37N\sdot0.15m}{0.034m}=163.24N;$

Answer:$F_2=163.24N.$