Problem 9 - Master Resource Book in JEE Main Physics

Question

A body of mass \(3 \mathrm{~kg}\) acted upon by a constant force is displaced by \(s\) metre, given by relation \(s=\frac{1}{3} t^{2}\), where \(t\) is in second. Work done by the force in \(2 \mathrm{~s}\) (a) \(\frac{8}{3} \mathrm{~J}\) (b) \(\frac{19}{5} \mathrm{~J}\) (c) \(\frac{5}{19} \mathrm{~J}\) (d) \(\frac{3}{8} \mathrm{~J}\)

Step-by-Step Solution

Answer

The work done by the force in 2 seconds is \(\frac{8}{3} \text{ J}\).

Step 1: Find the displacement expression

The displacement of the body is given by the equation \( s = \frac{1}{3}t^2 \). This equation tells us how the displacement \( s \) changes with time \( t \).

Step 2: Calculate the displacement at \( t = 2 \) seconds

Substitute \( t = 2 \) into the displacement equation: \[ s = \frac{1}{3}(2)^2 = \frac{1}{3}(4) = \frac{4}{3} \text{ meters} \] Thus, the displacement at \( t = 2 \) seconds is \( \frac{4}{3} \) meters.

Step 3: Find the velocity expression by differentiating the displacement

Velocity \( v(t) \) is the derivative of displacement \( s(t) \) with respect to time:\[ v(t) = \frac{d}{dt}\left( \frac{1}{3}t^2 \right) = \frac{2}{3}t \]

Step 4: Calculate the velocity at \( t = 2 \) seconds

Substitute \( t = 2 \) into the velocity equation:\[ v(2) = \frac{2}{3}(2) = \frac{4}{3} \text{ m/s} \] So, the velocity at \( t = 2 \) seconds is \( \frac{4}{3} \text{ m/s} \).

Step 5: Find the acceleration by differentiating the velocity

Acceleration \( a(t) \) is the derivative of velocity \( v(t) \) with respect to time:\[ a(t) = \frac{d}{dt}\left( \frac{2}{3}t \right) = \frac{2}{3} \text{ m/s}^2 \] Since it is a constant force, the acceleration is constant at \( \frac{2}{3} \text{ m/s}^2 \).

Step 6: Apply Newton's Second Law to find the force

According to Newton's Second Law:\[ F = ma \]Substituting the given mass \( m = 3 \text{ kg} \) and calculated acceleration \( a = \frac{2}{3} \text{ m/s}^2 \) gives:\[ F = 3 \times \frac{2}{3} = 2 \text{ N} \] So, the force acting on the body is \( 2 \text{ N} \).

Step 7: Calculate the work done by the force in 2 seconds

Work done \( W \) by a constant force is given by the formula:\[ W = Fs \]Using the force \( F = 2 \text{ N} \) and displacement \( s = \frac{4}{3} \text{ meters} \), we find:\[ W = 2 \times \frac{4}{3} = \frac{8}{3} \text{ J} \] Thus, the work done by the force in 2 seconds is \( \frac{8}{3} \text{ J} \).

Key Concepts

Displacement-Time Relationship

When we talk about displacement-time relationships in physics, it's all about understanding how the position of an object changes over time. Here, the given displacement equation is \( s = \frac{1}{3}t^2 \). This means the displacement \( s \) depends on the square of time \( t \).
In simpler terms, if you graph this equation with displacement on the y-axis and time on the x-axis, you will get a parabolic curve opening upwards. This indicates that the further out in time you get, the more the displacement increases.

• Time \( t \) is the independent variable, meaning it's what you choose or change, like setting a stopwatch.
• Displacement \( s \) is the dependent variable, meaning it changes in response to changing time.
• The coefficient \( \frac{1}{3} \) gives the rate at which displacement increases over time.

Understanding this relationship helps us predict how far an object moves in a given time, which is crucial in solving physics problems.

Velocity Calculation in Physics

Velocity tells us how fast something is moving and in which direction. To find the velocity from a displacement-time relationship, we differentiate the displacement equation with respect to time. In this case, the displacement equation is \( s = \frac{1}{3}t^2 \).
Differentiating gives us \( v(t) = \frac{d}{dt}(\frac{1}{3}t^2) = \frac{2}{3}t \). This formula means that the velocity \( v \) increases linearly with time.

• At any specific moment, substitute the time value into the velocity equation to find the speed.
• The velocity here is expressed in meters per second (m/s), showing rate and direction.
• For \( t = 2 \) seconds, the velocity is \( \frac{4}{3} \text{ m/s} \).

Velocity is crucial because it helps determine how quickly an object will reach a certain point.

Acceleration and Newton's Second Law

Acceleration indicates how the object’s velocity changes with time. In our exercise, acceleration is found by differentiating the velocity function \( v(t) = \frac{2}{3}t \) with respect to time, resulting in a constant acceleration \( a(t) = \frac{2}{3} \text{ m/s}^2 \). This means that the object gains \( \frac{2}{3} \) meters per second of speed every second, which is steady due to the constant force.
According to Newton’s Second Law, \( F = ma \), acceleration helps us find the force needed to change the motion of the body. Here,

• The mass \( m \) is \( 3 \text{ kg} \).
• The acceleration \( a \) is \( \frac{2}{3} \text{ m/s}^2 \).

Plug these into Newton's Second Law to find the force: \( F = 3 \times \frac{2}{3} = 2 \text{ N} \). This law shows the direct proportional relationship between force and the product of mass and acceleration.

Work-Energy Principle

Work in physics refers to the energy transferred to or from an object via the application of a force along a displacement. The work-energy principle states that the work done on an object is equal to the change in its kinetic energy.
The formula for work done \( W \) is \( W = Fs \). Here, the force \( F \) is \( 2 \text{ N} \) and the displacement \( s \) over 2 seconds is \( \frac{4}{3} \text{ meters} \). Therefore, the work done is:
\[ W = 2 \times \frac{4}{3} = \frac{8}{3} \text{ J} \]

• Work is calculated in joules \( \text{J} \).
• It considers only the component of the force in the direction of displacement.
• The calculated work indicates the amount of energy transferred as a result of the force applied over \( \frac{4}{3} \) meters.

This principle is fundamental in physics, linking the concepts of force, energy, and motion.