Question

A light elastic cord of length $l_0 = 5.0$ m and stiffness $k = 50$ N/m is inside a fixed box with a small portion of length $x_0 = 20$ cm left out of a hole on a wall of the box as shown in the figure. If the cord is pulled out of the box, a frictional force $F = 25$ N acts on the cord. How much work has to be done in slowly pulling out the entire length of the cord?

Answer 1

744 J

Answer 2

The problem asks for the total work done in slowly pulling out an elastic cord from a box. The total work done is the sum of the work done against the elastic force of the cord and the work done against the frictional force.

Given parameters:

• Initial length of cord outside the box, $x_0 = 20$ cm = 0.20 m
• Total length of the cord, $l_0 = 5.0$ m
• Stiffness of the cord, $k = 50$ N/m
• Frictional force, $F = 25$ N

1. Work done against the elastic force ($W_{elastic}$):

The stiffness $k$ implies that the elastic force exerted by the cord when it is stretched by a length $x$ from its unstretched state is given by Hooke's Law, $F_{elastic} = kx$. In this context, $x$ represents the length of the cord that has been pulled out of the box, assuming that the cord starts to stretch as soon as it is pulled out.
The cord is initially pulled out by $x_0$ and is finally pulled out completely, meaning the length outside becomes $l_0$.

The work done against the elastic force is given by the integral of the elastic force over the displacement:

$W_{elastic} = \int_{x_0}^{l_0} F_{elastic} dx = \int_{x_0}^{l_0} kx dx$

$W_{elastic} = k \left[ \frac{x^2}{2} \right]_{x_0}^{l_0} = \frac{1}{2} k (l_0^2 - x_0^2)$

Substitute the given values:

$W_{elastic} = \frac{1}{2} (50 \text{ N/m}) ((5.0 \text{ m})^2 - (0.20 \text{ m})^2)$

$W_{elastic} = 25 \text{ N/m} (25 \text{ m}^2 - 0.04 \text{ m}^2)$

$W_{elastic} = 25 \text{ N/m} (24.96 \text{ m}^2)$

$W_{elastic} = 624 \text{ J}$

2. Work done against the frictional force ($W_{friction}$):

A constant frictional force $F = 25$ N acts on the cord as it is pulled out. The cord is pulled out from an initial length $x_0$ outside to a final length $l_0$ outside.

The total displacement over which the frictional force acts is $\Delta x = l_0 - x_0$.

The work done against friction is:

$W_{friction} = F \times (l_0 - x_0)$

$W_{friction} = 25 \text{ N} \times (5.0 \text{ m} - 0.20 \text{ m})$

$W_{friction} = 25 \text{ N} \times (4.8 \text{ m})$

$W_{friction} = 120 \text{ J}$

3. Total work done ($W_{total}$):

The total work done in slowly pulling out the entire length of the cord is the sum of the work done against the elastic force and the work done against the frictional force.

$W_{total} = W_{elastic} + W_{friction}$

$W_{total} = 624 \text{ J} + 120 \text{ J}$

$W_{total} = 744 \text{ J}$