Suppose that 6 J of work is needed to stretch a spring from its natural length of 26 cm to a length of 39 cm. (a) How much work (in J) is needed to stretch the spring from 30 cm to 35 cm

Workdone = 0.89 Joules

Explanation:

Given the following data;

Workdone = 6J

Extension = 39 - 26 = 13cm to meters = 13/100 = 0.13m

The workdone to stretch a string is given by the formula;

Workdone = ½ke²

Where;

k is the constant of elasticity.

e is the extension of the string.

We would solve for string constant, k;

6 = ½*k*0.13²

6 = ½*k*0.0169

Cross-multiplying, we have;

12 = 0.0169k

k = 12/0.0169

k = 710.06

a. To find the workdone when e = 30, 35.

Extension = 35 - 30 = 5 to meters = 5/100 = 0.05m

Workdone = ½*710.06*0.05²

Workdone = 355.03*0.0025

Workdone = 0.89 Joules

Therefore, the amount of work (in J) needed to stretch the spring from 30 cm to 35 cm is 0.89.

nah.   i'm just sleepy.

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