Question

If it requires 3.0 J of work to stretch a particular spring by 2.1 cm from its equilibrium length, how much more work will be required to stretch it an additional 4.1 cm ?

Answer 1

Answer:

Work done = 13605.44

Explanation:

Data provided in the question:

For elongation of 2.1 cm (0.021 m) work done by the spring is 3.0 J

The relation between Energy (U) and the elongation (s) is given as:

U = $\frac{1}{2}kx^2$   ................(1)

where,

k is the spring constant

on substituting the valeus in the above equation, we get

3.0 = $\frac{1}{2}k\times0.021^2$

or

k = 13605.44 N/m

now

for the elongation x = 2.1 + 4.1 = 6.2 cm = 0.062 m

using the equation 1, we have

U = $\frac{1}{2}\times13605.44\times (0.062)^2$

or

U = 26.149 J

Also,

Work done = change in energy

or

W = 26.149 - 3.0 = 23.149 J