When a force of 50 N is applied, a spring is compressed by 10 cm. Find the potential energy of the spring when it is compressed by 20 cm.

Question

Question

Answer 1

CONCEPT:

Spring force: In an ideal spring, the force required to stretch a string from its equilibrium position is directly proportional to the extension of the spring.

This is known as Hooke's law for springs:

Fs = -k x

Where Fs is the spring force, x is the displacement from the equilibrium position and k is the spring constant.

\(PE = {1\over 2}kx^2\)

where k is the spring constant, x is the elongation or compression in the spring.

CALCULATION:

Given that F_s = 50 N then x = 10 cm = 0.1 m

Fs = -kx

50 = k × 0.1

k = 500 N/m

Now potential energy when compressed by 20 cm, i.e. 0.2 m will be:

\(PE = {1\over 2}kx^2 = {1\over 2}\times 500 \times (0.2)^2\)

PE = 10 J

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