Answer:
(1) 30 g
(2) [tex]3.5316 \times 10^{-3}[/tex] Joules
Explanation:
The unstretched length of the spring =30 cm
Assuming that the spring is linear.
(1) Let the mass of the sand be m kg. So, the gravitational force acting on the sand in the downward direction is mg N, which acts on the spring.
Due to the sand load, the length of the spring becomes 45 cm, so the extension in the spring, [tex]x_0[/tex] = 45-30=15 cm = 0.15 m.
As the spring is linear, so
[tex]mg=kx_0\cdots(i)[/tex]
where k is spring constant and g is the acceleration due to gravity.
When the 20 g mass is placed on the top of the sand, then the total load on the spring = m+0.02 kg
The gravitational force acting on the spring due to this load [tex]= (m+0.02)g[/tex] N
As the length of the spring becomes 55 cm, so the extension in the spring, [tex]x_1[/tex] = 55-30=25 cm = 0.25 m.
Now, [tex](m+0.02)g=kx_1\cdots(ii)[/tex]
Dividing equation (ii) by (i), we have
[tex]\frac{(m+0.02)g}{mg}=\frac{kx_1}{kx_0} \\\\\frac{(m+0.02)}{m}=\frac{x_1}{x_0} \\\\1+\frac{0.02}{m}=\frac{x_1}{x_0} \\\\\frac{0.02}{m}=\frac{x_1}{x_0}-1 \\\\[/tex]
On putting all the values, we have
[tex]\frac{0.02}{m}=\frac{0.25}{0.15}-1 \\\\\frac{0.02}{m}=\frac{2}{3} \\\\m=\frac{3}{2}\times 0.02 \\\\[/tex]
m=0.03 kg = 30 g
Therefore, the mass of the sand is 30 g.
(2) Putting m=0.03 kg in the equation (i) to get the value of spring constant, k.
(0.03)g=k(0.15)
0.03 x 9.81 = k(0.15) [as g=9.81 m/[tex]s^2[/tex]]
k=(0.03 x 9.81)/0.15
k=1.962 N/m
The spring is compressed from its original length to the length of 24cm, so the magnitude of change in length, x=30-24=6cm=0.06m.
So, the work done in compressing the spring,
[tex]w= \frac 1 2 k x^2[/tex]
[tex]w=\frac 1 2 \times 1.962 \times 0.06^2[/tex]
[tex]w=3.5316 \times 10^{-3}[/tex] Joules.
Therefore the work done in compressing the spring is [tex]3.5316 \times 10^{-3}[/tex] Joules.
