The force required to stretch a hooke’s-law spring varies from 0 n to 20.2 n as we stretch the spring by moving one end 5.39 cm from its unstressed position. find the force constant of the spring. answer in units of n/m. 014 (part 2 of 2) 10.0 points find the work done in stretching the spring.

The force constant of the spring is 374.77N/m

The workdone in stretching the spring is 0.54439Joules

According to Hooke's law, the force applied to the spring is directly proportional to its extension. This is expressed as:

F = ke

k is the spring constant/Force constant

e is the extension

F is the applied force

Given the following parameters

F = 20.2N

e = 5.39cm = 0.0539m

Get the force constant

k = F/e

k = 20.2/0.0539

k = 374.77N/m

The force constant of the spring is 374.77N/m

Next is to get the workdone on the spring

Work done = 1/2ke²

W = 1/2 * 374.77 * 0.0539²

W = 0.54439

Hence the workdone in stretching the spring is 0.54439Joules

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Explanation:

Given

Force changes from 0 N to 20.2 N

Extension in spring is x=5.39 cm

we know Force applied to stretch spring is given by

(b)Work Done to Stretch it

less than one you can't create energy

multiply 2kg by latent heat of vap