Answer:
Work done = 87.5 J
Explanation:
Given:
Force required to stretch the spring (F) = 250 N
Extension of the spring (x) = 30 cm = 0.30 m [1 cm = 0.01 m]
So, spring constant (k) of the spring is given as:
[tex]k=\frac{F}{x}\\\\k=\frac{250\ N}{0.30\ m}=833.33\ N/m[/tex]
Also given:
Initial length of the spring (x₁) = 20 cm = 0.20 m
Final length of the spring (x₂) = 50 cm = 0.50 m
Now, work done in stretching the spring from an initial length (x₁) to final length (x₂) is given as:
[tex]W=\frac{1}{2}k(x_2^2-x_1^2)[/tex]
Plug in the given values and solve for 'W. This gives,
[tex]W=\frac{1}{2}\times \frac{250}{0.3}\times (0.50^2-0.20^2)\\\\W=\frac{1250}{3}\times (0.25-0.04)\\\\W=\frac{1250\times 0.21}{3}=87.5\ J[/tex]
Therefore, work is done in stretching the spring from 20 centimeters to 50 centimeters is 87.5 J
