Answer:
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
[tex]F=kx[/tex]
[tex]20.2=k\times 5.39\times 10^{-2}[/tex]
[tex]k=374.76 N/m[/tex]
(b)Work Done to Stretch it
[tex]W=\frac{1}{2}kx^2[/tex]
[tex]W=0.5kx^2[/tex]
[tex]W=0.5\times 374.76\times (0.0539)^2[/tex]
[tex]W=0.544 J[/tex]
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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