A spring with spring constant of 28 N/m is stretched 0.2 m from its equilibrium position.

How much work must be done to stretch it
an additional 0.06 m?
Answer in units of J

To find the amount of work required to stretch the spring an additional 0.06 m, we can use the formula for the work done on a spring:

Work = (1/2)k(x^2)

Where:
- Work is the amount of work done on the spring (in Joules, J)
- k is the spring constant (in Newton per meter, N/m)
- x is the displacement from the equilibrium position (in meters)

Given that the spring constant is 28 N/m and the initial displacement is 0.2 m, we can calculate the initial amount of work done:

Initial Work = (1/2)(28 N/m)(0.2 m)^2
= 0.56 J

Now, we need to calculate the final amount of work when the spring is stretched an additional 0.06 m.

Final displacement = Initial displacement + Additional displacement
= 0.2 m + 0.06 m
= 0.26 m

Final Work = (1/2)(28 N/m)(0.26 m)^2
= 0.728 J

Therefore, the amount of work required to stretch the spring an additional 0.06 m is 0.728 Joules (J).

F1 = (0.2m/1m) * 28N = 5.6 N. To stretch

spring 0.2 m.
Work = F1 * d = 5.6 * 0.2 - 1.12 J.

F2 = (0.2m+0.06m)/1m) * 28 N.=7.28 N.
Work = F2 * d = 7.28 * 0.26 = 1.89 J.

1.89-1.12 = 0.77 J. to stretch it an additional 0.06 m.