Question

A cable in the form of a spiral roll, as shown in the figure, has a linear density $0.25\,kg{m^{ - 1}}$. It is uncoiled at a uniform speed of $2\,m{s^{ - 1}}$. If the total length of the cable is $15\,m$, the work done in uncoiling is (neglect friction) :

Answer 1

The Work done by a body in moving with a uniform velocity is equal to the kinetic energy of the body. Find the kinetic energy of the spiral roll to find the work done.The SI unit of work is the joule (J), which is defined as the work done by a force of 1 Newton in moving an object through a distance of 1 meter in the direction of the force.

Formula used:
The kinetic energy of a body is given by the formula,
$K.E = \dfrac{1}{2}m{v^2}$
where, $K.E$ is the kinetic energy of the body $m$ is the total mass of the body and $v$ is the velocity of the body.

Complete step by step answer:
We have given here a spiral roll which is uncoiling at a uniform speed of $2\,m{s^{ - 1}}$ which has a length of $15m$ and the linear density of mass is $0.25\,kg{m^{ - 1}}$.Now, we have to find the work done by the body. Now, we can treat this system as a system with point mass. If the roll was a point mass that would still be uncoiling with the same speed.

Then the work done by the roll will be the change in kinetic energy of the particle. Now, here the coil has a linear mass density of $0.25\,kg{m^{ - 1}}$ and a length of $15\,m$so the total mass of the roll is, $0.25 \times 15 = 3.75kg$. The uncoiling velocity of the coil is $2\,m{s^{ - 1}}$. Now, the coil was initially at rest hence initial kinetic energy is zero. So, the work done by the coil will be,
$\text{Work done} = \dfrac{1}{2} \times 3.75 \times {2^2}\,J$
$\therefore \text{Work done}= 7.5\,J$
Hence, work done in uncoiling is $7.5\,J$ .

Hence, Option B is the correct answer.

Note: If it was given to find the energy required in coiling the spiral roll we can find its potential energy using the formula for potential energy of a spring which is given by, $P.E = \dfrac{1}{2}K{x^2}$ where $K$ is the spring constant which depends on the property of the spring and $x$ is the displacement of the spring.