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Question

If the resultant force acting on a particle is perpendicular to its velocity then-

1. The speed of the particle changes.

2. The speed of the particle does not change.

3. Kinetic energy of the particle does not change.

Choose the correct code.

The correct answer is

Only 2 and 3

Understanding Force Perpendicular to Velocity in Physics

When a resultant force acts on a particle, its effect on the particle's motion depends on the direction of the force relative to the particle's velocity. A particularly important case is when the resultant force is perpendicular to the particle's instantaneous velocity.

Work Done by a Perpendicular Force

The work done by a force (\(\vec{F}\)) during a small displacement (\(d\vec{r}\)) is given by the dot product \(dW = \vec{F} \cdot d\vec{r}\). The instantaneous displacement \(d\vec{r}\) is always in the direction of the instantaneous velocity (\(\vec{v}\)). Therefore, if the force \(\vec{F}\) is perpendicular to the velocity \(\vec{v}\), it is also perpendicular to the displacement \(d\vec{r}\).

In this case, the angle between \(\vec{F}\) and \(d\vec{r}\) is 90 degrees. The work done is:

\[ dW = |\vec{F}| |d\vec{r}| \cos(90^\circ) = |\vec{F}| |d\vec{r}| \times 0 = 0 \]

The total work done by the resultant force over any path will be the sum of these infinitesimal work elements, which is also zero.

\[ W = \int dW = \int 0 = 0 \]

Work-Energy Theorem and Kinetic Energy

The work-energy theorem states that the change in kinetic energy (\(\Delta KE\)) of a particle is equal to the total work done (\(W\)) on it by the resultant force.

\[ \Delta KE = W \]

Since the work done by the resultant force is zero when it is perpendicular to the velocity, the change in kinetic energy is also zero:

\[ \Delta KE = 0 \]

This means that the kinetic energy of the particle remains constant.

\[ KE = \text{constant} \]

Kinetic Energy and Speed

Kinetic energy is defined as \(KE = \frac{1}{2}mv^2\), where \(m\) is the mass of the particle and \(v\) is its speed. For a given particle, the mass \(m\) is constant. If the kinetic energy \(KE\) is constant, then \(\frac{1}{2}mv^2\) must be constant. Since \(\frac{1}{2}m\) is a non-zero constant, \(v^2\) must also be constant.

\[ v^2 = \frac{2KE}{m} = \text{constant} \]

If \(v^2\) is constant, then the speed \(v\) must also be constant (assuming \(v \ge 0\)).

Analyzing the Statements

Let's examine each statement based on our understanding:

  • Statement 1: The speed of the particle changes.

    As we derived, if the resultant force is perpendicular to the velocity, the speed of the particle remains constant. Therefore, this statement is incorrect.

  • Statement 2: The speed of the particle does not change.

    Our analysis shows that the speed remains constant under these conditions. Therefore, this statement is correct.

  • Statement 3: Kinetic energy of the particle does not change.

    Since the work done by the perpendicular force is zero, the change in kinetic energy is zero, meaning the kinetic energy remains constant. Therefore, this statement is correct.

Based on the analysis, statements 2 and 3 are correct.

Statement Analysis Correct/Incorrect
1. The speed of the particle changes. Work done by force \(\perp\) velocity is zero. Zero work means constant KE. Constant KE means constant speed. Incorrect
2. The speed of the particle does not change. Constant KE implies constant speed. Correct
3. Kinetic energy of the particle does not change. Work done by force \(\perp\) velocity is zero. Work-energy theorem says \(\Delta KE = W\). So \(\Delta KE = 0\). Correct

Conclusion on Particle Motion

When the resultant force on a particle is always perpendicular to its velocity, the force only changes the direction of the velocity, not its magnitude (speed). This type of force is often called a centripetal force, which causes the particle to move along a curved path, such as circular motion.

Revision Table: Force & Velocity Relationship

Condition Work Done Change in KE Change in Speed Effect on Motion
Force \(\perp\) Velocity Zero (\(W=0\)) Zero (\(\Delta KE=0\)) Zero (\(\Delta v=0\)) Changes direction only (e.g., circular motion)
Force || Velocity (same direction) Positive Positive Positive (speed increases) Changes magnitude and direction (if initial motion is not in line)
Force || Velocity (opposite direction) Negative Negative Negative (speed decreases) Changes magnitude and direction (if initial motion is not in line)

Additional Information on Force and Energy

The relationship between force, velocity, work, and energy is fundamental in physics, particularly in mechanics. Here are some related concepts:

  • Power: The rate at which work is done, \(P = \frac{dW}{dt} = \vec{F} \cdot \vec{v}\). When force is perpendicular to velocity, the instantaneous power is zero, meaning the force is not adding or removing energy from the particle's motion.
  • Conservative Forces: Forces for which the work done is path-independent (e.g., gravity, electrostatic force). Work done by a conservative force results in a change in potential energy.
  • Non-Conservative Forces: Forces for which the work done is path-dependent (e.g., friction, air resistance). Work done by non-conservative forces results in a change in mechanical energy (sum of kinetic and potential energy).
  • Centripetal Force: A force that acts on an object moving in a circular path and is directed towards the center of the circle. This force is always perpendicular to the instantaneous velocity of the object, causing it to change direction but not speed (in uniform circular motion).

Understanding how the direction of the resultant force affects work and energy is crucial for solving various problems in dynamics and energy conservation.

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Important Questions from Newton's Laws of Motion

  1. Weight and mass of an object are defined with Newton’s laws of motion. Which among the following is true ?

  2. Which one of the following is not a contact force?

  3. If an object moves at a non-zero constant acceleration for a certain interval of time, then the distance it covers in that time

  4. A rigid body of mass 2 kg is dropped from a stationary balloon kept at a height of 50 m from the ground. The speed of the body when it just touches the ground and the total energy

    when it is dropped from the balloon are respectively

    (acceleration due to gravity = 9·8 m/s -2 )
  5. A body has a free fall from a height of 20 m. After falling through a distance of 5 m, the body would

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