Question

A spring requires a force of 25 n to produce an extension of 2 cm in it. what is the work done in further extending it by 3 cm

Answer 1

Answer: The work done in further extending spring by 3 cm is 0.5625 Joules.

Explanation:

Force exerted on spring,F = 25 N

Distance up to which spring stretched ,x= 2 cm = 0.02 m

(1 m = 100 cm)

$F=kx$

$k=\frac{F}{x}=\frac{25 N}{0.02 m}=1250 N/m$

Work done while further stretching the spring by distance x' = 3 cm.

Work done = $\frac{1}{2}k\times x'^2=\frac{1}{2}\times 1250 N/m\times (3\times 10^{-2} m)^2=0.5625 N m=0.5625 Joules$

The work done in further extending spring by 3 cm is 0.5625 Joules.

Answer 2

The force of a spring can be calculated by the expression:

F = kx  where k is a constant and x is the distance.

From the problem statement, we can solve the value of k from the initial conditions given as F = 25 N and x = 2 cm.

25 = k (0.02)
k = 1250

We use the work formula which is force times the distance resulting to the expression:

W = (1/2)(kx^2)
W = (1/2)(1250)(0.03^2)
W = 0.5625 J