If a force of 6 lbs is required to hold a spring stretched 5 inches beyond its natural length, then 38.4 lb-in. of work is done in stretching it from its natural length to 8 in. beyond its natural length.

Question

Quizplus · Accepted Answer

Work done = force applied × distance moved in the direction of force
From the given information,
Force required to stretch the spring by 5 inches = 6 lbs
Distance moved in the direction of force = 5 inches
Therefore, work done = 6 lbs × 5 inches = 30 lb-in.

To stretch the spring from its natural length to 8 inches beyond its natural length, we need to apply an additional force. Let's say the force required is f lbs.
The total force required to stretch the spring to 8 inches beyond its natural length is 6 lbs + f lbs.
The distance moved in the direction of force is 8 - 5 = 3 inches.
We know that the total work done is 38.4 lb-in.
Therefore,
38.4 = (6 + f) × 3
38.4 = 18 + 3f
3f = 38.4 - 18
3f = 20.4
f = 6.8 lbs

The work done in stretching the spring by an additional 3 inches (from 5 inches to 8 inches) is:
force applied × distance moved in the direction of force
= 6.8 lbs × 3 inches
= 20.4 lb-in.

Therefore, the total work done in stretching the spring from its natural length to 8 inches beyond its natural length is:
30 + 20.4 = 50.4 lb-in.

Hence, the given statement is true.

If a Force of 6 Lbs Is Required to Hold