Question

Plzzz help me solve this .

A vertical spring of unstretched length 30cm is clamped at its upper end. when sand is placed in a pan attached to the lower end of the spring its length becomes 45cm .when 20g mass is placed on top of the sand the length increases to 55cm
(1) determine the mass of the sand
(2) if the spring in a above is compressed from its original length to the length of 24cm,calculatethe work done in compressing the spring​

Answer 1

(1) 30 g

(2)
`3.5316 * 10^(-3)` Joules

Step-by-step 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,
`x_0` = 45-30=15 cm = 0.15 m.

As the spring is linear, so

`mg=kx_0\cdots(i)`

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
`= (m+0.02)g` N

As the length of the spring becomes 55 cm, so the extension in the spring,
`x_1` = 55-30=25 cm = 0.25 m.

Now,
`(m+0.02)g=kx_1\cdots(ii)`

Dividing equation (ii) by (i), we have

`((m+0.02)g)/(mg)=(kx_1)/(kx_0) \\\\((m+0.02))/(m)=(x_1)/(x_0) \\\\1+(0.02)/(m)=(x_1)/(x_0) \\\\(0.02)/(m)=(x_1)/(x_0)-1 \\\\`

On putting all the values, we have

`(0.02)/(m)=(0.25)/(0.15)-1 \\\\(0.02)/(m)=(2)/(3) \\\\m=(3)/(2)* 0.02 \\\\`

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/
`s^2`]

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,

`w= \frac 1 2 k x^2`

`w=\frac 1 2 * 1.962 * 0.06^2`

`w=3.5316 * 10^(-3)` Joules.

Therefore the work done in compressing the spring is
`3.5316 * 10^(-3)` Joules.