Problem 10 - A Complete Resource Book in Physics for JEE Main

Question

A person brings a mass of \(1 \mathrm{~kg}\) from infinity to a point \(A\). Initially the mass was at rest but it moves at a speed of \(2 \mathrm{~m} / \mathrm{s}\) as it reaches \(A\). The work done by the person on the mass is \(-3 \mathrm{~J}\). The potential at \(A\) is (A) \(-3 \mathrm{~J} / \mathrm{kg}\) (B) \(-2 \mathrm{~J} / \mathrm{kg}\) (C) \(-5 \mathrm{~J} / \mathrm{kg}\) (D) None of these

Step-by-Step Solution

Answer

The potential at point A is (A) \(-3\ \mathrm{J/kg}\).

Step 1: Identify relevant variables

We are given: - Mass, m = 1 kg - Initial speed, v_initial = 0 m/s (at rest) - Final speed, v_final = 2 m/s - Work done by the person, W = -3 J We need to find the potential at point A (V_A).

Step 2: Apply the Work-Energy Theorem

The Work-Energy Theorem states that the work done by all forces acting on an object is equal to the change in kinetic energy of the object. \( W = \Delta KE \) where ΔKE is the change in kinetic energy.

Step 3: Calculate the change in kinetic energy

The kinetic energy is given by the formula: \( KE = \frac{1}{2}mv^2 \) where - m is the mass of the object, and - v is its velocity. The change in kinetic energy is the difference between the final and initial kinetic energy: \( \Delta KE = KE_{final} - KE_{initial} \) We plug in the given values: \( \Delta KE = \frac{1}{2}(1\ \mathrm{kg})(2\ \mathrm{m/s})^2 - \frac{1}{2}(1\ \mathrm{kg})(0\ \mathrm{m/s})^2 = 2\ \mathrm{J} - 0\ \mathrm{J} = 2\ \mathrm{J} \)

Step 4: Calculate the change in potential

Since the work done equals the change in kinetic energy, we can find the change in potential by equating the work done to the change in potential energy: \( W = \Delta PE \) We plug in the given values: \( -3\ \mathrm{J} = \Delta PE \)

Step 5: Calculate the potential at point A

Now we can find the potential at point A: \( V_A = \frac{\Delta PE}{m} \) We plug in the values: \( V_A = \frac{-3\ \mathrm{J}}{1\ \mathrm{kg}} = -3\ \mathrm{J/kg} \) Therefore, the potential at point A is: (A) \(-3\ \mathrm{J/kg}\).