Question

A spring with spring constant of 34 N/m is stretched 0.12 m from its equilibrium position. How much work must be done to stretch it an additional 0.062 m

Answer 1

The amount of work required to stretch the spring an additional 0.062 m is approximately 0.561 J.

Step-by-step explanation:

To calculate the work required to stretch the spring an additional 0.062 m, we can use the formula for work done by a spring force, which is given by W = ½k(x2² - x1²), where W is the work done, k is the spring constant, and x2 and x1 are the final and initial displacements, respectively. In this case, the spring constant is 34 N/m and the initial displacement is 0.12 m. We can assume that the spring is initially at its equilibrium position, so x1 = 0. Using these values, we can calculate the work as follows:

W = ½(34 N/m)((0.12+0.062)² - 0.12²)
= ½(34 N/m)(0.182² - 0.012²)
= ½(34 N/m)(0.033124 - 0.000144)
= ½(34 N/m)( 0.03298)
≈ 0.5606 N·m
≈ 0.561 J

Therefore, the amount of work required to stretch the spring an additional 0.062 m is approximately 0.561 J.

Answer 2

Answer:0.253Joules

Step-by-step explanation:

First, we will calculate the force required to stretch the string. According to Hooke's law, the force applied to an elastic material or string is directly proportional to its extension.

F = ke where;

F is the force

k is spring constant = 34N/m

e is the extension = 0.12m

F = 34× 0.12 = 4.08N

To get work done,

Work is said to be done if the force applied to an object cause the body to move a distance from its initial position.

Work done = Force × Distance

Since F = 4.08m, distance = 0.062m

Work done = 4.08 × 0.062

Work done = 0.253Joules

Therefore, work done to stretch the string to an additional 0.062 m distance is 0.253Joules