A door spring is difficult to stretch a What maximum force d

A door spring is difficult to stretch. (a) What maximum force do you need to exert on a relaxed spring with a 1.2 times 10^4-N/m spring constant to stretch it 6.0 cm from its equilibrium position? (b) How much does the elastic potential energy of the spring change? (c) Determine its change in elastic potential energy as it returns from the 6.0-cm stretch position to a 3.0-cm stretch position, (d) Determine its elastic potential energy change as it moves from the 3.0-cm stretch position back to its equilibrium position.

Solution

Solution

(a) If say \'x\' is the displacement from equilibrium position and \'k\' is the spring constant then Force that spring will exert is defined by Hooke\'s law

F = -k*x (1)

Then for 6 cm stretch, Force that need to be exerted will be

|F| = 1.2 x 10⁴ N/m * 6.0 cm (2)

Keeping the dimensions from \'cm\' to \'m\' will give

|F| = 1.2 x 10⁴ N/m * 0.06 (3)

|F| = 720 N (4)

b) Elastic potential energy (U₁) of the spring for displacement \'x\' can be given as

U₁ = 1/2*k*x² (5)

U₁ = (1/2)*(1.2 * 10⁴)*(0.06)² (6)

U₁ = 21.6 J (7)

(c) Potential energy stored in spring when it was stretched to 3 cm (U₂)

U₂ = (1/2)*(1.2*10⁴)*(0.03)² (8)

U₂ = 5.4 J (9)

Change in potential energy as it returns from 6.0 cm stretch to 3.0 cm stretch

U₁ - U₂ = 21.6 J - 5.4 J (10)

U₁ - U₂ = 16.2 J (11)

(d) At equilibrium position x = 0

U₃ = 0 (12)

Hence change in potential energy when spring is stretched from 3.0 cm to its equilibrium position

U₂ - U₃ = 5.4 J (13)