A spring with spring constant 18 N/m has a length of 10 cm. It is pulled to a length of 16 cm. What is the work required to do this?
According to Hooke’s Law, the force required to increase the length of a spring is proportional to the length by which the spring has been pulled and is given by the relation as F = kx, where k is the spring constant and x is the change from the normal length. In this problem the spring has a spring constant of 18N/m.
When the spring is 10 cm long we require a force of 0 N to pull it. This gradually increases as the spring is pulled further by an infinitesimal length dL and at 16 cm it is equal to 18*.06 N.
To find the total work done in pulling the spring from 10 cm to 16 cm, we have to find the definite integral of kx for x = 0 to x = 6.
Work required is Int [kx], x= 0 to x = .06m
=> kx^2/2, x = 0 to x = 0.06
=> 18*x^2/2, x = 0 to x = 0.06
=> 9*x^2, x = 0 to x = 0.06
At x = 0, this is equal to 0.
At x = 0.06, this is equal to 9*0.06^2 = 0.0324
Subtracting the value at x = 0 from that at x = 6 cm, we get 0.0324 J.
Therefore the work required to pull the spring from 10 cm to 16 cm is 0.0324J.