Answer :
(a) 0.714 cm
First of all, we need to find the spring constant of the spring. This can be done by using Hooke's law:
[tex]F=kx[/tex]
where
F is the force applied on the spring
k is the spring constant
x is the stretching of the spring
At the beginning, the force applied is the weight of the block of m = 4.20 kg hanging on the spring, therefore:
[tex]F=mg=(4.20)(9.8)=41.2 N[/tex]
The stretching of the spring due to this force is
x = 2.00 cm = 0.02 m
Therefore, the spring constant is
[tex]k=\frac{F}{x}=\frac{41.2}{0.02}=2060 N/m[/tex]
Now, a new object of 1.50 kg is hanging on the spring instead of the previous one. So, the weight of this object is
[tex]F=mg=(1.50)(9.8)=14.7 N[/tex]
And so, the stretching of th spring in this case is
[tex]x=\frac{F}{k}=\frac{14.7}{2060}=0.00714 m = 0.714 cm[/tex]
(b) 1.65 J
The work done on a spring is given by:
[tex]W=\frac{1}{2}kx^2[/tex]
where
k is the spring constant
x is the stretching of the spring
In this situation,
k = 2060 N/m
x = 4.00 cm = 0.04 m is the stretching due to the external agent
So, the work done is
[tex]W=\frac{1}{2}(2060)(0.04)^2=1.65 J[/tex]