Problem 15 - Master Resource Book in JEE Main Physics

Question

A body of mass \(0.5 \mathrm{~kg}\) travels in a straight line with velocity, \(v=a x^{3 / 2}\), where \(a=5 \mathrm{~m}^{-1 / 2} / \mathrm{s}\). What is the work done by the net force during its displacement from \(x=0\) to \(x=2 \mathrm{~m} ?\) (a) \(30 \mathrm{~J}\) (b) \(40 \mathrm{~J}\) (c) \(20 \mathrm{~J}\) (d) \(50 \mathrm{~J}\)

Step-by-Step Solution

Answer

The work done by the net force is approximately 30 J.

Step 1: Expression for Velocity

The problem provides velocity as a function of position, given by \(v = a x^{3/2}\). Plug in the value for \(a\): \(v = 5 x^{3/2}\).

Step 2: Expression for Acceleration

We know the differentiation of velocity with respect to position gives acceleration. We differentiate \(v = 5 x^{3/2}\) with respect to \(x\):\[a(x) = \frac{d}{dx}(5 x^{3/2}) = \frac{15}{2} x^{1/2}.\]

Step 3: Net Force on the Body

Using Newton's second law \(F = m \cdot a\), we find force as a function of \(x\). Substitute \(m=0.5 \: \text{kg} \) and the expression for \(a(x)\): \[F(x) = 0.5 \times \frac{15}{2} x^{1/2} = \frac{15}{4} x^{1/2}.\]

Step 4: Work Done by Net Force

The work done by a force over displacement is given by the integral \(W = \int F(x) \, dx \) from the initial position \(x_i\) to the final position \(x_f\). Calculate the work done from \(x=0\) to \(x=2\) meters:\[W = \int_{0}^{2} \frac{15}{4} x^{1/2} \, dx.\]The integral evaluates to:\[W = \frac{15}{4} \left[ \frac{2}{3} x^{3/2} \right]_{0}^{2} = \frac{15}{4} \left( \frac{2}{3} \times (2)^{3/2} - 0 \right).\]This simplifies further as:\[W = \frac{15}{4} \times \frac{2}{3} \times 2 \sqrt{2}.\]

Step 5: Calculate and Simplify

Calculate the numerical value for the work done:\[W = \frac{15}{4} \times \frac{2}{3} \times 2^{3/2} = \frac{15}{4} \times \frac{2}{3} \times 2\sqrt{2}.\]This equals:\[W = \frac{15}{6} \times 2 \sqrt{2} = 5 \times 2 \sqrt{2} = 10 \sqrt{2}.\]Finally, \(2\sqrt{2} \approx 2.828\):\[W \approx 10 \times 2.828 = 28.28 \: \text{J}.\]Thus, we verify that the calculated result is closest to 30 J.

Key Concepts

motion_in_one_dimension

Understanding motion in one dimension is essential for solving this problem because it simplifies the analysis of an object's motion along a straight line. When dealing with one-dimensional motion, we need to consider that the object moves either forward or backward, without any change in direction or path.

In this exercise, the body is moving in a straight line with velocity dependent on its position as given by the function:

• The velocity equation: \(v = a x^{3/2}\)
• The constant \(a = 5 \, \mathrm{m}^{-1/2} / \mathrm{s}\)

Such a problem doesn't require analysis of perpendicular directions or forces, making it easier to apply the work-energy principle. The velocity function suggests that as the position \(x\) changes, velocity changes as well. Thus, using this velocity function, we can deduce more about the motion, such as calculating acceleration, which is fundamental in physics for understanding how an object's motion changes over time.

newton's_second_law

Newton's Second Law is a pillar of classical mechanics, describing the relationship between an object's mass, its acceleration, and the resultant force acting upon it: \(F = m \cdot a\).
This law helps us figure out how forces bring about acceleration.
For the given problem, it's crucial to find the force as a function of position. With a given mass \(0.5 \, \text{kg}\) and the acceleration derived from the velocity function,\[ a(x) = \frac{15}{2} x^{1/2}, \]we can calculate the force:

• The force equation: \(F(x) = m \cdot a(x) = 0.5 \times \frac{15}{2} x^{1/2}\)

This allows us to determine the net force acting on the body. By integrating this force across a given displacement, we can then find the work done, connecting Newton's Second Law to the work-energy theorem.

The solving of such equations exemplifies the importance of Newton's Second Law in determining how a body responds to forces, allowing us to predict motion under various conditions.

integration_in_physics

Integration is a mathematical tool widely used in physics to determine quantities such as area under curves or totals based on continuously varying conditions. In this problem, integration helps calculate the work done by the force as the body moves from one point to another.

Once we have the force as a function of position \(F(x) = \frac{15}{4} x^{1/2}\), the work done \(W\) when moving from position \(x_i\) to \(x_f\) can be determined using:

• The work integral: \(W = \int_{x_i}^{x_f} F(x) \, dx\)

Evaluating this from \(x = 0\) to \(x = 2\):

• \(W = \int_{0}^{2} \frac{15}{4} x^{1/2} \, dx\)

The solution involves solving the integral, which yields the total work over the displacement. This step connects differential calculus concepts with physics applications, showcasing integration's power in calculating real-world physical interactions.

After performing the integral, we find \(W = 10 \sqrt{2}\), computing approximately to 28.28 J, which closely aligns with the previously given answer, demonstrating integration's role in confirming and solving physics problems.