Chapter 6, Problem 1P

(a)

To determine

Whether the statement is true or false.

Explanation of Solution

Introduction:

The work energy theorem states that the net or total work done on a particle is equal to the change in its kinetic energy.

According to the work energy theorem, the net or total work done on a particle is equal to the change in its kinetic energy.

Write the expression for the work-energy theorem.

  $W\text{ = }\frac{1}{2}\left(m{v}_f^2-m{v}_i^2\right)$

Here, $W$ is the work done on particle, $m$ is the mass of the particle, $v_f$ and $v_i$ are the final and initial velocities of the particle respectively.

Conclusion:

Thus, for non-zero work done, the initial and final speeds mustdiffer i.e. there must be a change in speed of the particle. Hence, the given statement is true.

(b)

To determine

Whether the statement is true or false.

Explanation of Solution

According to the work energy theorem, the net or total work done on a particle is equal to the change in its kinetic energy.

Write the expression for the work-energy theorem.

  $W\text{ = }\frac{1}{2}\left(m{v}_f^2-m{v}_i^2\right)$

Here, $W$ is the work done on particle, $m$ is the mass of the particle, $v_f$ and $v_i$ are the final and initial velocities of the particle respectively.

Introduction:The work energy theorem states that the net or total work done on a particle is equal to the change in its kinetic energy.

Conclusion:Thus, for non-zero work done, the initial and final velocities must differ i.e. there must be a change in velocity of the particle. Hence, the given statement is true.

(c)

To determine

Whether the statement is true or false.

Explanation of Solution

Introduction:

The work energy theorem states that the net or total work done on a particle is equal to the change in its kinetic energy.

According to the work energy theorem, the net or total work done on a particle is equal to the change in its kinetic energy.

Write the expression for the work-energy theorem.

  $W\text{ = }\frac{1}{2}\left(m{v}_f^2-m{v}_i^2\right)$

Here, $W$ is the work done on particle, $m$ is the mass of the particle, $v_f$ and $v_i$ are the final and initial velocities of the particle respectively.

For a particle moving in straight line, i.e. its direction of motion is not changing, if its speed changes with time then, the net work done on it will be non-zero while if its speed is constant then, the net work done on it will be zero.

Conclusion:

Thus, for a particle moving in straight line, i.e. its direction of motion is not changing, and its speed changing with time then, the net work done on it will be non-zero. Hence, the given statement is true.

(d)

To determine

Whether the statement is true or false.

Explanation of Solution

Introduction:

The work energy theorem states that the net or total work done on a particle is equal to the change in its kinetic energy.

According to the work energy theorem, the net or total work done on a particle is equal to the change in its kinetic energy.

Write the expression for the work-energy theorem.

  $W\text{ = }\frac{1}{2}\left(m{v}_f^2-m{v}_i^2\right)$

Here, $W$ is the work done on particle, $m$ is the mass of the particle, $v_f$ and $v_i$ are the final and initial velocities of the particle respectively.

For a particle at rest, the speed is zero throughout and thus, work done on it will be zero.

Conclusion:

Thus, for a particle at constant rest, the initial and final velocities of the particle are same and thus work done is zero.

(e)

To determine

Whether the statement is true or false.

Explanation of Solution

Introduction:

The work done on a particle is given by the dot product of force acting on it and its displacement.

For a particle experiencing a force in a particular direction and having a displacement, the work done on it is given by the dot product of force acting on it and its displacement.

If the angle between force and displacement is ${90}^{\circ }$ i.e. the force is acting perpendicular to the motion of the particle, the work done on it is always zero.

Conclusion:

Thus, if a force is always perpendicular to the velocity of particle, the angle betweenforce and displacement is ${90}^{\circ }$ and no work is done on the particle. Hence, the given statement is true.