A spring in equilibrium is 12 centimeters long and an external force of 60 dynes compresses the spring to a length of 10 centimeters. The work done by the spring in compressing it from equilibrium to 8 centimeters, in dyne-centimeters, is

A)-264
B)-240
C)-238
D)-212
E)-200

Question

A spring in equilibrium is 12 centimeters long and an external force of 60 dynes compresses the spring to a length of 10 centimeters. The work done by the spring in compressing it from equilibrium to 8 centimeters, in dyne-centimeters, is
​
A)-264
B)-240
C)-238
D)-212
E)-200

Quizplus · Accepted Answer

The work done by the spring is calculated using Hooke's Law and the formula for work done by a spring: W = 1/2 * k * (x1^2 - x2^2). First, find the spring constant k using F = k * x, where F = 60 dynes and x = 2 cm (12 cm - 10 cm), giving k = 30 dynes/cm. Then, calculate the work done from 12 cm to 8 cm: W = 1/2 * 30 * (4^2 - 0^2) = 240 dyne-cm.

A Spring in Equilibrium Is 12 Centimeters Long and an External