Chapter 7, Problem 1P

To determine

The correct statement.

Explanation of Solution

Given:

Mass of two cylinders is not equal and pulley is frictionless.

Formula used:

Write the expression for kinetic energy.

  $K=\frac{1}{2}m{v}^2$

Here, $m$ is the mass and $v$ is the velocity of an object or system.

Write the expression for the change in the kinetic energy of the system.

  $\Delta K=K_f-K_i$

Since the system is released from rest initial speed of the system is zero.

Substitute $0$ for $v_i$ in above expression.

  $\begin{array}{c}\Delta K=K_f-0\\ =K_f\end{array}$

Therefore, $\Delta K>0$ .

Write the expression for potential energy.

  $U=mgh$

Here, $g$ is the acceleration due to gravity and $h$ is height.

Write the expression for the change in the potential energy of the system.

  $\Delta U=U_f-U_i$

Here, $\Delta U$ is the change in potential energy, $U_f$ is the final potential energy of the system and $U_i$ is the initial potential energy of the system.

Consider $m_1$ is the mass of the first cylinder and $m_2$ is the mass of the second cylinder. Since $m_2>m_1$ and heavier one is going downwards so potential energy of the system is decreasing and change in potential energy is less than zero $\Delta U<0$ since the mechanical energy of the system will remain conserved.

Conclusion:

Since the change in potential energy is less than zero and change in kinetic energy is greater than zero. Thus, option (a) is correct.

Since the height of the system is changing so change in potential energy cannot be zero. Thus, option (b) is incorrect.

The system is moving downwards with some speed so the change in kinetic energy cannot be zero. Thus, option (c) is incorrect.

Since option (b) and (c) are incorrect. Thus, option (d) is incorrect.

Change in kinetic energy could not be less than zero. Thus, option (e) is incorrect.