Question

A spring on a horizontal surface can be stretched and held 0.5 m from its equilibrium position with a force of 60 N. a. How much work is done in stretching the spring 5.5 m from its equilibrium position? b. How much work is done in compressing the spring 1.5 m from its equilibrium position?

Answer 1

Answer:

a)1815Joules b) 185Joules

Explanation:

Hooke's law states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically;

F = ke where;

F is the applied force

k is the elastic constant

e is the extension of the material

From the formula, k = F/e

F1/e1 = F2/e2

If a force of 60N causes an extension of 0.5m of the string from its equilibrium position, the elastic constant of the spring will be ;

k = 60/0.5

k = 120N/m

a) To get the work done in stretching the spring 5.5m from its position,

Work done by the spring = 1/2ke²

Given k = 120N/m, e = 5.5m

Work done = 1/2×120×5.5²

Work done = 60× 5.5²

Work done = 1815Joules

b) work done in compressing the spring 1.5m from its equilibrium position will be gotten using the same formula;

Work done = 1/2ke²

Work done =1/2× 120×1.5²

Works done = 60×1.5²

Work done = 135Joules

Answer 2

Answer:

a) W = 1815 J

b) W = 135 J

Explanation:

We need to model the force using Hooke's law

We first get the elastic constant, k

A force of 60 N causes an extension of 0.5 m

F = 60 N, e = 0.5 m

F = k * e

60 = 0.5 k

k = 60/0.5

k = 120 N/m

Therefore the force in terms of the extension, x is:

F(x) = 120x

a) Work done in stretching the spring 5.5 m from its equilibrium position

b = 0 m, a = 5.5 m

$W = \int\limits^a_b {F(x)} \, dx$

$W = \int\limits^a_b {120x} \, dx \\a =0, b = 5.5$

$W = 60x^{2} \\W = 60 [5.5^{2} -0^{2} ]$

$W = 1815 J$

b) Work required to compress the spring 1.5 m from its equilibrium position

$W = \int\limits^a_b {F(x)} \, dx$ b = 0, a = -1.5

$W = 60x^{2} \\W = 60 [(-1.5)^{2} -0^{2} ]$

W = 135 J

The work done to compress the spring 1.5 m from its equilibrium position is 135 J