Question

A uniform rope of mass $m$ and length $l_0$ lies straight on a rough horizontal table with a portion hanging as shown in the figure. If the hanging portion is increased beyond one fourth of its length the rope begins to slide. Initially one fourth of the length of the rope is hanging, and the rope is in equilibrium. The length of the hanging portion is increased slightly, and the rope is released as a result the rope begins to slide. Find work done by frictional forces, work done by gravity, and work done by the force of normal reaction of the table on the rope.

Answer 1

Work done by frictional forces = $-\frac{3mgl_0}{32}$, Work done by gravity = $\frac{15mgl_0}{32}$, Work done by normal reaction = 0

Answer 2

The problem involves a uniform rope sliding on a rough horizontal table. We need to calculate the work done by friction, gravity, and the normal reaction force.

1. Determine the coefficient of kinetic friction ($\mu_k$):

Initially, one-fourth of the rope's length ($l_0/4$) is hanging, and the rope is in equilibrium. This means the tension due to the hanging part is balanced by the maximum static friction force on the part lying on the table.

Let the mass per unit length of the rope be $\lambda = m/l_0$.

Mass of the hanging portion = $\lambda (l_0/4) = m/4$.

Force due to gravity on the hanging portion (tension) = $(m/4)g$.

Length of the rope on the table = $l_0 - l_0/4 = 3l_0/4$.

Mass of the rope on the table = $\lambda (3l_0/4) = 3m/4$.

Normal force exerted by the table on the rope = $N = (3m/4)g$.

Maximum static friction force = $f_s^{max} = \mu_s N = \mu_s (3m/4)g$.

At equilibrium:

$(m/4)g = \mu_s (3m/4)g$

$1/4 = \mu_s (3/4)$

$\mu_s = 1/3$.

Since the rope begins to slide, we assume the coefficient of kinetic friction $\mu_k$ is equal to the coefficient of static friction $\mu_s$, so $\mu_k = 1/3$.

2. Work done by the normal reaction force ($W_N$):

The normal force exerted by the table on the rope is always perpendicular to the displacement of the rope along the table. Therefore, the work done by the normal reaction force is zero.

$W_N = 0$

3. Work done by gravity ($W_g$):

The rope starts with $l_0/4$ hanging and slides until the entire length $l_0$ is hanging.

Let $x$ be the length of the hanging portion. The force of gravity pulling the rope down is due to the mass of the hanging portion, which is $F_g(x) = (\lambda x) g = (m/l_0) x g$.

The rope moves from an initial hanging length $x_i = l_0/4$ to a final hanging length $x_f = l_0$.

Work done by gravity is the integral of the gravitational force over the displacement:

$W_g = \int_{x_i}^{x_f} F_g(x) dx = \int_{l_0/4}^{l_0} \frac{mg}{l_0} x dx$

$W_g = \frac{mg}{l_0} \left[ \frac{x^2}{2} \right]_{l_0/4}^{l_0}$

$W_g = \frac{mg}{l_0} \left( \frac{l_0^2}{2} - \frac{(l_0/4)^2}{2} \right)$

$W_g = \frac{mg}{l_0} \left( \frac{l_0^2}{2} - \frac{l_0^2}{32} \right)$

$W_g = \frac{mg}{l_0} \left( \frac{16l_0^2 - l_0^2}{32} \right) = \frac{mg}{l_0} \left( \frac{15l_0^2}{32} \right)$

$W_g = \frac{15mgl_0}{32}$

4. Work done by frictional forces ($W_f$):

The kinetic friction force acts on the portion of the rope lying on the table.

When the hanging length is $x$, the length of the rope on the table is $(l_0 - x)$.

The mass of the rope on the table is $\lambda (l_0 - x) = (m/l_0)(l_0 - x)$.

The normal force on the table is $N(x) = (m/l_0)(l_0 - x)g$.

The kinetic friction force is $f_k(x) = \mu_k N(x) = \frac{1}{3} \frac{mg}{l_0} (l_0 - x)$.

The friction force opposes the motion, so the work done by friction is negative.

$W_f = \int_{x_i}^{x_f} -f_k(x) dx = \int_{l_0/4}^{l_0} -\frac{1}{3} \frac{mg}{l_0} (l_0 - x) dx$

$W_f = -\frac{mg}{3l_0} \int_{l_0/4}^{l_0} (l_0 - x) dx$

$W_f = -\frac{mg}{3l_0} \left[ l_0 x - \frac{x^2}{2} \right]_{l_0/4}^{l_0}$

$W_f = -\frac{mg}{3l_0} \left[ (l_0(l_0) - \frac{l_0^2}{2}) - (l_0(l_0/4) - \frac{(l_0/4)^2}{2}) \right]$

$W_f = -\frac{mg}{3l_0} \left[ (l_0^2 - \frac{l_0^2}{2}) - (\frac{l_0^2}{4} - \frac{l_0^2}{32}) \right]$

$W_f = -\frac{mg}{3l_0} \left[ \frac{l_0^2}{2} - (\frac{8l_0^2 - l_0^2}{32}) \right]$

$W_f = -\frac{mg}{3l_0} \left[ \frac{l_0^2}{2} - \frac{7l_0^2}{32} \right]$

$W_f = -\frac{mg}{3l_0} \left[ \frac{16l_0^2 - 7l_0^2}{32} \right] = -\frac{mg}{3l_0} \left[ \frac{9l_0^2}{32} \right]$

$W_f = -\frac{3mgl_0}{32}$

The final answers are:

Work done by frictional forces: $-\frac{3mgl_0}{32}$

Work done by gravity: $\frac{15mgl_0}{32}$

Work done by the force of normal reaction of the table on the rope: $0$

The question asks for these three values.