Question

A rubber band has a "spring constant" of 68 N/m. When the rubber band is pulled by a force of 13 N, how far will it stretch and what is its potential energy?

Answer 1

GIVEN:

• Spring constant of rubber, k = 68 N/m
• Force applied, F = 13 N

TO FIND:

• Stretching length, x
• Potential energy, U

FORMULAE:

• F = - kx

• U = kx²/2

SOLUTION:

STEP 1: To find stretching length

$F = - kx \\ \\ 13 = - 68 \times x \\ \\ x = - \frac{13}{68} \\ \\ x = - 0.191 \: m \\ \\ x = - 0.191 \times 100 \: cm \\ \\ x = - 19.1 \: cm$

Stretching length is 0.191 m or 19.1 cm.

Negative sign indicates that the direction of force is opposite to the direction of elongation.

STEP 2: To find potential energy

$U = \dfrac{1}{2} k {x}^{2} \\ \\ U = \dfrac{1}{2} \times 68 \times {(0.191)}^{2} \\ \\ U = \dfrac{1}{2} \times 68 \times { (\dfrac{13}{68} )}^{2} \\ \\ U = \dfrac{1}{2} \times 68 \times \dfrac{13}{68} \times \dfrac{13}{68} \\ \\ U = \dfrac{1}{2} \times \dfrac{13 \times 13}{68} \\ \\ U = \dfrac{1}{2} \times \dfrac{169}{68} \\ \\ U = \dfrac{1}{2} \times 2.49 \\ \\ U = 1.245 \: joule$

The potential energy is 1.245 J.

ANSWER:

• Stretching length is 0.191 m or 19.1 cm.

• The potential energy is 1.245 J.

OTHER FORMULA:

The potential energy stored in the rubber band can also be given by the formula,

U = F²/2k or U = Fx/2