Question

A force of 10 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work W is done in stretching it from its natural length to 9 in. beyond its natural length

Answer 1

11.47J

Step-by-step explanation:

Hooke's law states that the force or load (F) applied to an elastic material (e.g a spring) is directly proportional to the extension or compression (e) caused by the load. i.e

F ∝ e

F = k e ---------------(i)

where;

k = proportionality constant called the spring or elastic constant.

From the question,

F = Force = 10 lb

[Remember that 1 lb = 4.45 N]

=> F = 10 x 4.45 N = 44.5 N

Also;

e = extension = 4 in

[Remember that 1 in = 0.0254 m]

e = 4 x 0.0254 m = 0.1016 m

Substitute these values into equation (i) to get the spring constant of the spring as follows;

44.5 = k x 0.1016

k =
`(44.5)/(0.1016)`

k = 439N/m

Now, lets calculate the work done

The workdone W, in stretching a spring from its natural length to some length beyond the natural length, is given by;

W =
`(1)/(2)` x k x e² --------------------(ii)

Where;

k = the spring's constant = 439N/m [as calculated above]

e = extension = 9 in = 9 x 0.0254m = 0.2286m [Recall: 1 in = 0.0254m]

Substitute these values into equation (ii) as follows;

W =
`(1)/(2)` x 439 x (0.02286)²

W = 11.47 Nm or 11.47J

Therefore, the work done is 11.47J

Answer 2

114.44 J

Step-by-step explanation:

From Hook's Law,

F = ke................. Equation 1

Where F = Force required to stretch the spring, k = spring constant, e = extension.

make k the subject of the equation

k = F/e.............. Equation 2

Given: F = 10 lb = (10×4.45) N = 44.5 N, e = 4 in = (4×0.254) = 1.016 m.

Substitute into equation 2

k = 44.5/1.016

k = 43.799 N/m

Work done in stretching the 9 in beyond its natural length

W = 1/2ke²................. Equation 3

Given: e = 9 in = (9×0.254) = 2.286 m, k = 43.799 N/m

Substitute into equation 3

W = 1/2×43.799×2.286²

W = 114.44 J