A certain spring that obeys Hooke's law is stretched by an external agent. The work done in stretching the spring by 10 cm is 4 J. How much additional work is required to stretch the spring an additional 10 cm?

12J

The work done (W) is stretching a spring that obeys Hooke's law is given by;

W  = k x²              -------------(i)

Where;

k = the spring constant measured in N/m

x = extension of the spring.

From the question;

When the spring is stretched (extended) by 10cm, the workdone is 4J.

This implies that;

At x = 10cm = 0.1m, W = 4J

Substitute these values into equation (i) to get the spring constant (k) of the spring as follows;

4 = * k (0.1)²

4 = * k (0.01)

4 = 0.005k

k = 4 / 0.005

k = 800N/m

Therefore, the spring constant of the spring is 800N/m

Now, let's get the work done in stretching the spring an additional 10cm.

This means that the total extension X, is the sum of the initial extension (10cm) and the current extension (10cm).

i.e

X = 10cm + 10cm = 20cm  = 0.2m

Substitute the value of X for x in equation (i) to get the work done as follows;

W = k X²

W = k (0.2)²

W = k (0.04)

W = 0.02k                ---------------(ii)

Where;

k = 800N/m      [as calculated above]

Substitute this value into equation (ii) as follows;

W = 0.02(800)

W = 16J

The work done to stretch the spring by a total extension of 20cm is 16J

Now the additional work required to stretch the spring 10cm further is given by the difference between the total work done and the initial work done as follows;

16 - 4 = 12J

Therefore, the additional work done is 12 J

absorption lines are where light has been absorbed by the atom thus you see a dip in the spectrum whereas.

emission spectra have spikes in the spectra due to atoms releasing photons at those wavelengths.

explanation: