It takes 5.06 J of work to stretch a Hooke's-law spring 13.3 cm from its unstressed length. How much the extra work is required to stretch it an additional 13.4 cm? Answer in units of J. It takes 5.06 J of work to stretch a Hooke's-law spring 13.3 cm from its unstressed length. How much the extra work is required to stretch it an additional 13.4 cm? Answer in units of J.

Answers:

extra work required = 0.038 J

Explanation:

let X = 13.3 cm and Y = 13.4 cm, W1 be the work done to stretch the spring 13.3 cm and W2 be the work done to stretch the spring 13.4cm. let F be the force stretching the spring.

from Work and Energy principles:

W1 = F*X*cos (Ф) = F*X, since the force is acting in the direction at which the spring is stretching, cos (Ф) = 1

then:

W1 = F*X

F = W1/X

= 5.06 / (13.3*10^-2)

= 38.05 J

and,

W2 = F*Y*cos (Ф) = F*Y, ince the force is acting in the direction at which the spring is stretching, cos (Ф) = 1

W2 = F*Y

= (38.5) * (13.4*10^*2)

= 5.098 J

therefore, the extra required is 5.098 - 5.06 = 0.038 J.