Question

A closed circular tube of average radius 15 cm , whose inner walls are rough, is kept in vertical plane. A block of mass 1 kg just fit inside the tube. The speed of block is 22 m s − 1 , when it is introduced at the top of tube. After completing five oscillations, the block stops at the bottom region of tube. The work done by the tube on the block is ________ J . (Given g = 10 m s^ − 2 ).

Answer 1

-245

Answer 2

The block starts at the top of the tube with a given initial speed and stops at the bottom of the tube. We can use the work-energy theorem. The total work done on the block is equal to the change in its kinetic energy. The forces acting on the block are gravity (conservative force) and the forces exerted by the tube (normal force and friction, which are non-conservative forces).

The work-energy theorem states: $W_{net} = \Delta KE$

The net work is the sum of the work done by gravity and the work done by the tube: $W_g + W_{\text{tube}} = KE_f - KE_i$

The work done by gravity is related to the change in potential energy: $W_g = -\Delta PE = -(PE_f - PE_i) = PE_i - PE_f$

Substituting this into the work-energy equation: $(PE_i - PE_f) + W_{\text{tube}} = KE_f - KE_i$ $W_{\text{tube}} = (KE_f - KE_i) + (PE_f - PE_i) = (KE_f + PE_f) - (KE_i + PE_i) = E_f - E_i$

where $E = KE + PE$ is the total mechanical energy. The work done by the tube is equal to the change in the total mechanical energy of the block.

Let's set the reference level for potential energy at the bottom of the tube ($h=0$). The radius of the tube is $R = 15$ cm = 0.15 m. The mass of the block is $m = 1$ kg. The acceleration due to gravity is $g = 10$ m/s$^2$.

Initial state (at the top): Height $h_i = 2R = 2 \times 0.15 = 0.3$ m. Initial speed $v_i = 22$ m/s. Initial kinetic energy $KE_i = \frac{1}{2}mv_i^2 = \frac{1}{2}(1 \text{ kg})(22 \text{ m/s})^2 = \frac{1}{2}(484) = 242$ J. Initial potential energy $PE_i = mgh_i = (1 \text{ kg})(10 \text{ m/s}^2)(0.3 \text{ m}) = 3$ J. Total initial mechanical energy $E_i = KE_i + PE_i = 242 \text{ J} + 3 \text{ J} = 245$ J.

Final state (at the bottom): Height $h_f = 0$ m. Final speed $v_f = 0$ m/s (the block stops). Final kinetic energy $KE_f = \frac{1}{2}mv_f^2 = \frac{1}{2}(1 \text{ kg})(0 \text{ m/s})^2 = 0$ J. Final potential energy $PE_f = mgh_f = (1 \text{ kg})(10 \text{ m/s}^2)(0 \text{ m}) = 0$ J. Total final mechanical energy $E_f = KE_f + PE_f = 0 \text{ J} + 0 \text{ J} = 0$ J.

The work done by the tube on the block is $W_{\text{tube}} = E_f - E_i$. $W_{\text{tube}} = 0 \text{ J} - 245 \text{ J} = -245$ J.

The work done by the tube on the block is the sum of the work done by the normal force and the work done by the frictional force. The normal force is always perpendicular to the displacement along the tube, so the work done by the normal force is zero. Therefore, the work done by the tube is equal to the work done by friction. The negative sign indicates that the work done by the tube (friction) is dissipative, removing energy from the mechanical system.

The information about "completing five oscillations" describes the path taken, which determines the total work done by friction. However, the work-energy theorem allows us to calculate the total work done by non-conservative forces based solely on the change in mechanical energy between the initial and final states, irrespective of the detailed path, as long as the initial and final states are correctly identified.