Question

The spring-flex exercise system consists of a spring with one end fixed and a handle on the other end. The idea is that you exercise your muscles by stretching the spring from its natural length, which is 33 cm. If a 190 Newton force is required to keep the spring stretched to a length of 43 cm, how much work is required to stretch it from 43 cm to 59 cm?

Answer 1

Calculating the work done to stretch a spring from 43 cm to 59 cm requires applying Hooke's Law to find the spring constant and then using the formula for work done on a spring, resulting in 24.32 Joules of work.

Step-by-step explanation:

To answer how much work is required to stretch the spring from 43 cm to 59 cm, it is important to apply Hooke's Law and the concept of elastic potential energy. Using the knowledge that a 190 Newton force is needed to hold the spring at 43 cm, we can derive the spring constant (k) since the spring's original length is 33 cm, giving us a displacement (Δx) of 10 cm, or 0.1 m. Hooke's Law states that F = kΔx, so k = F/Δx = 190N/0.1m = 1900 N/m. The work done W in stretching a spring is calculated using the formula W = 1/2kΔx², where Δx is the change in length from the initial to the final state. From 43 cm to 59 cm, the change in length is 16 cm or 0.16 m. Therefore, the work done in stretching the spring from 43 cm to 59 cm would be W = 1/2(1900N/m)(0.16m)² = 24.32 Joules.