Question

A particle of mass \(0.5 \mathrm{~kg}\) travels in a straight line with velocity \(\mathrm{v}=\mathrm{ax}^{3 / 2}\), Where \(\mathrm{a}=5 \mathrm{~m}^{[(-1) / 2]} \mathrm{s}^{-1}\). The work done by the net force during its displacement from \(\mathrm{x}=0\) to \(\mathrm{x}=2 \mathrm{~m}\) is (A) \(50 \mathrm{~J}\) (B) \(45 \mathrm{~J}\) (C) \(25 \mathrm{~J}\) (D) None of these

Short answer

The work done by the net force during the particle's displacement from \(x = 0 \mathrm{~m}\) to \(x = 2 \mathrm{~m}\) is \(50 \mathrm{~J}\). The correct answer is (A) \(50 \mathrm{~J}\).

Step-by-step solution

Find the initial and final velocities

First, evaluate the velocity function at the initial and final displacements: \(x = 0 \mathrm{~m}\) and \(x = 2 \mathrm{~m}\). Initial velocity (\(v_0\)): \[v_0 = a x^{3/2} = 5(0)^{3/2} = 0 \mathrm{~m/s}\] Final velocity (\(v_f\)): \[v_f = a x^{3/2} = 5(2)^{3/2} = 5(2\sqrt{2}) = 10\sqrt{2} \mathrm{~m/s}\]

Calculate the initial and final kinetic energies

Using the mass of the particle, we can calculate its initial and final kinetic energies, given by the formula \(K = \frac{1}{2}mv^2\). Initial kinetic energy (\(K_0\)): \[K_0 = \frac{1}{2}mv_0^2 = \frac{1}{2}(0.5 \mathrm{~kg})(0 \mathrm{~m/s})^2 = 0 \mathrm{~J}\] Final kinetic energy (\(K_f\)): \[K_f = \frac{1}{2}mv_f^2 = \frac{1}{2}(0.5 \mathrm{~kg})(10\sqrt{2} \mathrm{~m/s})^2 = 50 \mathrm{~J}\]

Calculate the work done using the work-energy theorem

According to the work-energy theorem, the work done (\(W\)) is equal to the change in kinetic energy, which is the difference between the final and initial kinetic energies: \[W = K_f - K_0 = 50 \mathrm{~J} - 0 \mathrm{~J} = 50 \mathrm{~J}\] Thus, the work done by the net force during the particle's displacement from \(x = 0 \mathrm{~m}\) to \(x = 2 \mathrm{~m}\) is \(50 \mathrm{~J}\). The correct answer is (A) \(50 \mathrm{~J}\).

Key concepts

Kinetic Energy

In physics, kinetic energy is the energy that an object possesses because of its motion. For a particle, kinetic energy (\(K\)) can be calculated using the formula:

• \[ K = \frac{1}{2} mv^2 \]

where \(m\) is the mass of the particle, and \(v\) is its velocity.

Kinetic energy represents how much work an object can do because of its speed and mass. The more mass and velocity, the more kinetic energy.

In the given problem, we initially calculate the kinetic energy at displacement \(x = 0\) and final displacement \(x = 2\) to find the work done through the change in this energy. When the velocity at the initial position is zero, the initial kinetic energy is naturally zero:

• \[ K_0 = \frac{1}{2}(0.5) imes 0^2 = 0 \, \text{J} \]

At the end of the path, using the final velocity, the kinetic energy is:

• \[ K_f = \frac{1}{2}(0.5)(10\sqrt{2})^2 = 50 \, \text{J} \]

So, the change in kinetic energy is crucial for determining the work done.

Velocity Function

A velocity function describes how the speed of a particle varies as it travels along a path. In this problem, the velocity function is given by \(v = ax^{3/2}\).This type of function is significant because:

• It relates the particle's position \(x\) to its velocity \(v\).
• In this scenario, \(a\) is a constant with units \(m^{(-1/2)}s^{-1}\).

To apply this function, substitute the specific values of \(x\) to find the velocity at each position:

• For the starting point \(x = 0\), \(v_0 = a(0)^{3/2} = 0 \, \text{m/s}\).
• For the end point \(x = 2\), \(v_f = a(2)^{3/2} = 10\sqrt{2} \, \text{m/s}\).

Understanding how velocity changes based on position can tell us a lot about the particle's movement, necessary for calculating kinetic energy and work done.

Work Done

Work done in physics refers to the energy transferred by a force acting upon an object as it moves a certain distance. It can be understood via the work-energy theorem, which states:

• The work done by the net force on an object is equal to the change in its kinetic energy.

The equation for work done (\(W\)) using the kinetic energy approach is:

• \[ W = K_f - K_0 \]

where \(K_f\) is the final kinetic energy, and \(K_0\) is the initial kinetic energy.

In this exercise, starting and final positions give us velocities that allow the calculation of kinetic energies. Therefore:

• From \(x = 0\) to \(x = 2\):
• \[ W = 50 \, \text{J} - 0 \, \text{J} = 50 \, \text{J} \]

This result means that while the particle was moving from \(x = 0\) to \(x = 2\), an energy of 50 Joules is transferred to it by the net force acting on it, and that is the amount of work done.