Question

Suppose that 4 J of work is needed to stretch a spring from its natural length of 36 cm to a length of 47 cm. (a) How much work is needed to stretch the spring from 41 cm to 45 cm? (Round your answer to two decimal places.) J (b) How far beyond its natural length will a force of 15 N keep the spring stretched? (Round your answer one decimal place.)

Answer 1

Answer:

Explanation:

Work done on a spring is expressed as $W = 1/2 ke^{2}$

k is the elastic constant

e is the extension of the material

If 4 J of work is needed to stretch a spring from its natural length of 36 cm to a length of 47 cm, then;

Work done = 4J and the extension e = 47 cm - 36 cm; e = 11 cm

11cm = 0.11m

Substituting the given values into the equation above to get the elastic constant;

$W = 1/2 ke^{2}\\4 = 1/2k(0.11)^{2} \\8 = 0.0121k\\k = 8/0.0121\\k = 661.16N/m$

a) In order to determine the amount of work needed work is needed to stretch the spring from 41 cm to 45 cm, wre will use the same formula as above.

$W = 1/2ke^{2} \\e = 0.45 - 0.41\\e = 0.04 m\\ k = 661.16N/m$

$W = 1/2 * 661.16 * 0.04^{2} \\W = 330.58*0.0016\\W = 0.53J (to\ 2d.p)$

b) According to hooke's law, F = ke where F is the applied force

We are to get the extension when a force of 15N is applied to the original length of the material.

e = F/k

e = 15/661.16

e = 0.02 m (to 1 d.p)

This means that the natural length of the spring will be stretched by 0.02 m when a force of 15N is applied to it.