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.

the second one and the forth one are in the purple because they are opinions and the rest are in the yellow

explanation:

the correct choice is a .

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