We can use Hooke's law to find the force constant of the spring:
F = kx
Where,
F: is the force applied to the spring,
k is the force constant of the spring,
x is the displacement of the spring from its equilibrium position
As the spring stretches 2.68 cm (or 0.0268 m) when a 3.00 kg object is suspended from it. The weight of the object is given by:
w = mg = 3.00 kg × 9.81 m/s² = 29.43 N
Which is the force F in the equation above so, to find k we can do the following:
k = F/x = w/x = 29.43 N / 0.0268 m = 1097 N/m
Hence, the force constant of the spring is 1097 N/m.
(b) Now it is simple as we already know the value for the force constant k. Again using the Hookes law
F = kx,
We already know k from above as the spring is the same we can use it again to find x.
w = mg = 1.50 kg × 9.81 m/s² = 14.72 N, which is F in Hookes's law.
F = kx
We can solve for x:
14.72 N = 1097 N/m × x
x = 0.0134 m = 1.34 cm
Hence, the spring stretches 1.34 cm when a 1.50 kg object is suspended from it.
(c) The work done to stretch the spring can be calculated using the following equation:
W = (1/2)kx²
where W is the work done,
k is the force constant of the spring,
x is the displacement of the spring from its equilibrium position.
We are given that x = 6.90 cm = 0.0690 m and k = 1097 N/m.
Using these values to solve for W we get
W = (1/2)(1097 N/m)(0.0690 m)²
W = 2.23 J
Hence, an external agent must do 2.23 J of work to stretch the spring 6.90 cm from its equilibrium, position.
