Problem 57 - Master Resource Book in JEE Main Physics

Question

A variable force given by the two-dimensional vector \(\mathbf{F}=\left(3 x^{2} \hat{\mathbf{i}}+4 \hat{\mathbf{j}}\right)\) acts on a particle. The force is in newton and \(X\) is in metre. What is the change in the kinetic energy of the particle as it moves from the point with coordinates \((2,3)\) to \((3,0)\) (The coordinates are in metres)? [AMU (Med) 2010] (a) \(-7]\) (b) zero (c) +7 J (d) \(+19 \mathrm{~J}\)

Step-by-Step Solution

Answer

The change in the kinetic energy is +7 J.

Step 1: Express Work Done as an Integral

The work done by the force \( \mathbf{F} \) as the particle moves from point A \((2,3)\) to point B \((3,0)\) is the line integral of the force along its path. This can be expressed as: \[ W = \int_{\mathbf{A}}^{\mathbf{B}} \mathbf{F} \cdot d\mathbf{r} \] where \( \mathbf{r} = (x, y) \) is the position vector.

Step 2: Parameterize the Path

Assume the particle follows a straight line path from \((2,3)\) to \((3,0)\). Parameterize this path using a parameter \( t \), such that: \[ x = 2 + (3-2)t = 2 + t \] \[ y = 3 + (0-3)t = 3 - 3t \] where \( 0 \leq t \leq 1 \).

Step 3: Calculate the Differential Element \(d\mathbf{r}\)

Differentiate the parametrization to find the differential of \( \mathbf{r} \). Thus, \[ dx = dt \] and \[ dy = -3dt \]. Therefore, \[ d\mathbf{r} = (dx, dy) = (\hat{\mathbf{i}} dt, -3 \hat{\mathbf{j}} dt) \].

Step 4: Substitute and Evaluate the Integral

Substitute \( \mathbf{F} \) and \( d\mathbf{r} \) into the integral for work done: \[ W = \int_0^1 (3(2+t)^2 \hat{\mathbf{i}} + 4 \hat{\mathbf{j}}) \cdot (\hat{\mathbf{i}} - 3 \hat{\mathbf{j}}) \, dt \] This simplifies to: \[ W = \int_0^1 \left[ 3(4 + 4t + t^2) - 12 \right] dt = \int_0^1 (12t + 3t^2) \, dt \].

Step 5: Perform the Integration

Solve the integral \[ \int_0^1 (12t + 3t^2) \, dt \]: \[= \left[ 6t^2 + t^3 \right]_0^1 \] \[ = \left(6(1)^2 + (1)^3\right) - \left(6(0)^2 + (0)^3\right) \] \[ = 6 + 1 = 7 \].

Step 6: Interpretation of Result

The work done by the force as the particle moves from \( (2,3) \) to \( (3,0) \) is equal to the change in the kinetic energy of the particle. Therefore, the change in kinetic energy is \(+7 \text{ J}\).

Key Concepts

Kinetic Energy

Kinetic energy is a measure of the energy an object possesses due to its motion. This concept is fundamental in physics, as it helps us understand how energy is transferred and transformed during the movement of particles. The formula for kinetic energy (\( KE \) ) is given by:\[ KE = \frac{1}{2} mv^2 \]where \( m \) is the mass of the object and \( v \) is its velocity. It is important to remember that kinetic energy is always a non-negative quantity, regardless of the direction of the object's movement. If an object's speed doubles, its kinetic energy increases by a factor of four.In many physical problems, like the one in the original exercise, we are interested in finding the change in kinetic energy as an object moves from one position to another. This change can be calculated by analyzing the work done by the forces acting on the object during its displacement. In the given problem, the line integral calculates the work done by force, which in turn is the change in the particle's kinetic energy.In summary:

• Kinetic energy depends on mass and velocity.
• The change in kinetic energy is the work done by the forces acting on the object.

Vector Calculus

Vector calculus is a branch of mathematics that deals with vector fields and operations on them. In this context, a vector field is a function that assigns a vector to every point in space. Understanding vector calculus is essential in physics, as it is often used to describe and analyze physical quantities that have direction and magnitude, like force or velocity fields.One important operation in vector calculus is the dot product, which combines two vectors to produce a scalar. In the original problem, the work done is calculated using a line integral, where the dot product of the force vector \( \mathbf{F} \) and an infinitesimal displacement \( d\mathbf{r} \) is integrated over a path.Key terms to know in vector calculus:

• Vectors: Quantities that have both magnitude and direction, denoted \( \hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}} \) for the unit vectors along the x, y, and z axes.
• Dot Product: A way to multiply two vectors to get a scalar, crucial in calculating work.
• Line Integrals: Used to calculate the work done by a vector field along a curve.

Understanding these concepts allows for the interpretation and manipulation of vector fields to solve real-world problems.

Line Integral

In mathematics, the line integral is a powerful tool used in calculus that accounts for the integration of a vector field along a curve. This concept is fundamental when calculating work done by a force that varies over the path traveled, as seen in the problem discussed.The line integral of a vector field \( \mathbf{F} \) over a curve \( C \) from point \( A \) to point \( B \) is defined as:\[ W = \int_{C} \mathbf{F} \cdot d\mathbf{r} \]Here, the dot denotes the dot product between the force and the infinitesimal element of the curve, represented as \( d\mathbf{r} \). This accounts for the force's influence along the path, considering its direction as well. The result is a scalar that gives the total work done by the vector field along that curve.
A line integral:

• Accumulates values over a curved path, rather than along an axis.
• Is essential in physics for calculating work done by forces.
• Requires parametrization of the path for evaluation purposes.

In the original exercise, the line integral was computed by first parametrizing the straight path between two points and then substituting this parametrization into the integral. This allowed for the calculation of work, revealing the change in kinetic energy of the particle—highlighting the practical application of line integrals in solving physics problems.