A spring in equilibrium is 4 meters long and an external force of 8 Newtons stretches the spring to a length of 4.5 meters. The work done by the spring in stretching it from 4.5 meters to 5 meters, in Newton-meters, is

A)-22
B)-17
C)-13
D)-9
E)-6

Question

A spring in equilibrium is 4 meters long and an external force of 8 Newtons stretches the spring to a length of 4.5 meters. The work done by the spring in stretching it from 4.5 meters to 5 meters, in Newton-meters, is
​
A)-22
B)-17
C)-13
D)-9
E)-6

Quizplus · Accepted Answer

The spring constant k can be found using Hooke's Law: F = kx, where F = 8 N and x = 0.5 m (4.5 m - 4 m). So, k = 8 / 0.5 = 16 N/m. The work done by the spring is given by the formula W = 0.5 * k * (x2^2 - x1^2), where x1 = 0.5 m and x2 = 1 m (5 m - 4 m). Substituting the values, W = 0.5 * 16 * (1^2 - 0.5^2) = 0.5 * 16 * (1 - 0.25) = 0.5 * 16 * 0.75 = 6 Nm.

A Spring in Equilibrium Is 4 Meters Long and an External