8. The force constant of a spring is 150. N/m. (a) How much force is required to stretch the spring 0.25 m? (b) How much work is done on the spring in that case?

(a)  The force required to stretch the spring is of 37.5 N.

(b)  The work done on the spring is of 4.6775 J.

Given data:

The force constant of a spring is, k = 150 N/m.

The stretching distance of spring is, x = 0.25 m.

(a)  

When some force is experienced on the spring, then due to this force then it will displace. Then, the expression for the force applied on the spring is given as,

Solve by substituting the values as,

Thus, we can conclude that the force required to stretch the spring is of 37.5 N.

(b)

And the expression for the work done on the spring is given as,

Solving as,

Thus, we can conclude that the work done on the spring is of 4.6775 J.

Learn more about the spring force here:

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a) 37.5N

b) 9.375Joules

Explanation:

a) According to Hooke's law

F = ke

k is the spring constant

e is the extension;

F = 150 * 0.25

F = 37.5N

b) Work done on the spring = 1/2ke^2

Work done on the spring = 1/2 * 150 * 0.25^2

Work done on the spring  = 75 * 0.0625

Work done on the spring  = 9.375Joules

double the distance = divide strength by 4

d. platinum would have the greatest temperature increase because it has the lowest specific heat.