Problem 47 - Master Resource Book in JEE Main Physics

Question

The potential energy of a particle of mass \(5 \mathrm{~kg}\) moving in the \(x y\)-plane is given by \(U=(-7 x+24 y) \mathrm{J}, x\) and \(y\) being in metre. If the particle starts from rest from origin, then speed of particle at \(t=2 \mathrm{~s}\) is (a) \(5 \mathrm{~ms}^{-1}\) (b) \(01 \mathrm{~ms}^{-1}\) (c) \(17.5 \mathrm{~ms}^{-1}\) (d) \(10 \mathrm{~ms}^{-1}\)

Step-by-Step Solution

Answer

Speed of the particle at \( t=2 \) s is \( 10 \text{ ms}^{-1} \) (Option d).

Step 1: Understand the Problem

We have a particle of mass 5 kg with potential energy given by the function \( U = -7x + 24y \) Joules, where \( x \) and \( y \) are in meters. We are asked to find the speed of the particle at \( t = 2 \) seconds if it starts from rest at the origin.

Step 2: Determine Forces from Potential Energy

The force on the particle can be found by taking the negative gradient of the potential energy. The force components are: \[ F_x = - \frac{\partial U}{\partial x} = 7 \text{ N} \]\[ F_y = - \frac{\partial U}{\partial y} = -24 \text{ N} \]

Step 3: Calculate the Net Force and Acceleration

The net force on the particle is given by the vector sum: \[ \vec{F} = (7 \hat{i} - 24 \hat{j}) \text{ N} \]Using Newton's Second Law, \( F = ma \), the acceleration \( \vec{a} \) is: \[ \vec{a} = \left(\frac{7}{5} \hat{i} - \frac{24}{5} \hat{j}\right) \text{ m/s}^2 \]

Step 4: Calculate the Velocity Components after 2 Seconds

Since the particle starts from rest, the initial velocity is \( \vec{v}_0 = 0 \).The velocity at time \( t \) is given by \( \vec{v} = \vec{v}_0 + \vec{a}t \).Substituting in the acceleration and \( t = 2 \): \[ v_x = \frac{7}{5} \times 2 = \frac{14}{5} \text{ m/s} \]\[ v_y = -\frac{24}{5} \times 2 = -\frac{48}{5} \text{ m/s} \]

Step 5: Calculate the Magnitude of Velocity (Speed)

The speed is the magnitude of the velocity vector:\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{\left(\frac{14}{5}\right)^2 + \left(-\frac{48}{5}\right)^2} \]\[ v = \sqrt{\frac{196}{25} + \frac{2304}{25}} = \sqrt{\frac{2500}{25}} = \sqrt{100} = 10 \text{ m/s} \]

Key Concepts

Force from Potential Energy

In physics, potential energy represents the stored energy within a system due to its position or configuration. This energy can be converted into kinetic energy, causing the object to move. To analyze particle motion in a potential energy field, we must first calculate the force exerted by this potential energy.
The force acting on a particle can be derived from the potential energy using the concept of the negative gradient. This involves finding the partial derivatives of the potential energy function with respect to each coordinate (in this case, x and y), and then changing their signs. For a potential energy function given by \( U = -7x + 24y \), the forces along the x and y directions are calculated as follows:

• \( F_x = - \frac{\partial U}{\partial x} = 7 \text{ N} \)
• \( F_y = - \frac{\partial U}{\partial y} = -24 \text{ N} \)

These force components tell us that the particle experiences a constant force of 7 N in the positive x-direction, and -24 N in the negative y-direction, affecting its motion. Understanding these force vectors is crucial for predicting how a particle will move in a potential field.

Newton's Second Law

Newton's Second Law of Motion is a key principle used to connect force and motion. It states that the force exerted on an object is equal to the mass of the object times its acceleration, expressed as \( F = ma \).
This relationship helps us find the acceleration of the particle we're studying. We already determined the net force acting on the particle as \( \vec{F} = (7 \hat{i} - 24 \hat{j}) \text{ N} \). With a particle mass of 5 kg, the acceleration in each direction can be computed by dividing each force component by the mass:

• \( a_x = \frac{7}{5} \text{ m/s}^2 \)
• \( a_y = -\frac{24}{5} \text{ m/s}^2 \)

The accelerations \(a_x\) and \(a_y\) reveal how quickly the particle speeds up or down along the x and y axes, offering insight into its overall motion. By understanding and applying Newton's Second Law, we set the groundwork for calculating velocity and determining the particle's path.

Particle Motion in Potential Field

Once we have the forces and accelerations, we can better understand how the particle moves in the potential field. The initial condition given is that the particle starts from rest at the origin, meaning the initial velocity is zero (\(\vec{v}_0=0\)).
In a potential field, the interplay of forces and initial conditions dictates the subsequent motion of the particle. The calculated accelerations indicate that initially, the particle will be pushed or pulled along the x and y axes, determined by the respective forces.
Since the particle starts from rest, it will begin to accelerate uniformly. The constant forces exerted will result in a linear increase in the velocity along both axes. This is described by the equation:\[ \vec{v} = \vec{v}_0 + \vec{a}t \]Where \(t\) is the time elapsed. As time progresses, the velocity vectors' magnitudes change, highlighting the importance of constant acceleration in predicting motion in potential fields.

Velocity Calculation

Velocity is a vector quantity representing the rate at which an object changes its position. To determine the velocity of the particle at \(t = 2 \text{ s}\), knowing it starts from rest, we utilize the previous step's acceleration values and apply them over the given time duration.
Given the velocities:

• \( v_x = \frac{7}{5} \times 2 = \frac{14}{5} \text{ m/s} \)
• \( v_y = -\frac{24}{5} \times 2 = -\frac{48}{5} \text{ m/s} \)

To find the speed (the magnitude of the velocity vector), compute:\[ v = \sqrt{v_x^2 + v_y^2} \]Substitute the known components:\[ v = \sqrt{\left(\frac{14}{5}\right)^2 + \left(-\frac{48}{5}\right)^2} \]After calculating, it results in:\[ v = \sqrt{100} = 10 \text{ m/s} \]This speed shows how quickly the particle moves away from its initial position at the specified time. The calculation reveals the effects of initial forces and acceleration, enabling us to predict the particle's behavior in its potential field.