Problem 18 - Master Resource Book in JEE Main Physics

Question

A force acts on a \(30 \mathrm{~g}\) particle in such a way that the position of the particle as function of time is given by \(x=3 t-4 t^{2}+t^{3}\), where \(x\) is in metre and \(t\) is in second. The work done during the first 4 seconds is (a) 5.28 J (b) \(450 \mathrm{~mJ}\) (c) \(490 \mathrm{~m} \mathrm{~J}\) (d) \(530 \mathrm{~mJ}\)

Step-by-Step Solution

Answer

The work done is 5.28 J (option a).

Step 1: Differentiate the position function

To find the force acting on the particle, calculate the velocity by differentiating the position function. The position function given is \( x = 3t - 4t^2 + t^3 \). \[ v(t) = \frac{dx}{dt} = 3 - 8t + 3t^2 \]

Step 2: Calculate velocity at specific times

Find the velocity at the beginning (\( t = 0 \)) and end (\( t = 4 \)) of the time interval. \[ v(0) = 3 - 8(0) + 3(0)^2 = 3 \, \text{m/s} \] \[ v(4) = 3 - 8(4) + 3(4)^2 = 3 - 32 + 48 = 19 \, \text{m/s} \]

Step 3: Use kinetic energy to find work done

The work done on the particle is the change in kinetic energy from \( t = 0 \) to \( t = 4 \). The expression for kinetic energy is \( KE = \frac{1}{2}mv^2 \). Given the mass \( m = 30 \, \text{g} = 0.03 \, \text{kg} \). Calculate initial and final kinetic energies. Initial kinetic energy (\( KE_0 \)): \[ KE_0 = \frac{1}{2} \times 0.03 \times (3)^2 = 0.5 \times 0.03 \times 9 = 0.135 \, \text{J} \]Final kinetic energy (\( KE_4 \)): \[ KE_4 = \frac{1}{2} \times 0.03 \times (19)^2 = 0.5 \times 0.03 \times 361 = 5.415 \, \text{J} \]

Step 4: Calculate the work done

Compute the work done over the time interval as the change in kinetic energy: \[ W = KE_4 - KE_0 = 5.415 \, \text{J} - 0.135 \, \text{J} = 5.28 \, \text{J} \]

Key Concepts

Kinetic Energy

Understanding kinetic energy allows us to determine how much work is done in a system. Kinetic energy is the energy a particle has due to its motion, and it's an essential part of physics, particularly in mechanics. - **Formula of Kinetic Energy**: The kinetic energy (KE) of a particle with mass \( m \) and velocity \( v \) is calculated using the formula: \[ KE = \frac{1}{2}mv^2 \] This equation shows that kinetic energy increases with the square of velocity. Hence, if an object's velocity doubles, its kinetic energy becomes four times greater. - **Relating Kinetic Energy to Work**: Work done on a particle results in a change in its kinetic energy. According to the work-energy theorem, the work done on an object is equal to the change in its kinetic energy \[ W = \Delta KE = KE_{ ext{final}} - KE_{ ext{initial}} \] By calculating the initial and final kinetic energies, we can find the work done on the particle.

Differentiation in Physics

Differentiation is a powerful tool in physics used to analyze changes in quantities. It provides insights into motion and forces acting on particles through functions of time and position.- **Velocity from Differentiation**: Differentiating the position function with respect to time gives the velocity function. For the specified problem, the position function was \( x = 3t - 4t^2 + t^3 \). By differentiating, we derive: \[ v(t) = \frac{dx}{dt} = 3 - 8t + 3t^2 \] This expression showcases how velocity changes with time. - **Acceleration from Velocity**: If needed, further differentiation of the velocity function provides acceleration, detailing how quickly the velocity changes over time.Differentiation thus acts as a bridge between different motion quantities, allowing us to transition from one concept, like position, to another, like velocity or acceleration.

Particle Motion

Particle motion entails understanding how particles move under the influence of forces, and it plays a significant role in mechanics.- **Position Function**: This describes the particle's location at any given time. In the given exercise, particle position as a function of time was described by \( x = 3t - 4t^2 + t^3 \), representing a curve that can be plotted on a graph.- **Velocity Function**: Obtained from differentiation, this tells us not just how fast the particle is moving, but also the direction of its motion. The velocity at certain times provides insight into the particle's speed and whether it is accelerating or decelerating.- **Influence of Forces**: Forces acting on particles cause changes in motion, akin to pushing or pulling. In our scenario, understanding how forces relate to changes in kinetic energy allows us to ascertain the work done.In essence, exploring particle motion includes examining how position, velocity, and forces interplay to describe a particle's journey over time. This motion is influenced by both inherent properties like mass and external factors such as forces applied.