Problem 70 - Master Resource Book in JEE Main Physics

Question

A force, \(\mathbf{F}=-K(y \hat{\mathbf{i}}+x \hat{\mathbf{j}})\) (where, \(K\) is a positive constant) acts on a particle moving in the \(x y\) plane. Starting from the origin, the particle is taken along the positive \(x\)-axis to the point \((a, 0)\) and then parallel to the \(y\)-axis to the point \((a, a)\). The total work done by the force, \(\mathbf{F}\) on the particle is (a) \(-2 \mathrm{Ka}^{2}\) (b) \(2 \mathrm{Ka}^{2}\) (c) \(-K a^{2}\) (d) \(K a^{2}\)

Step-by-Step Solution

Answer

The correct answer is (c) \(-Ka^2\).

Step 1: Determine Work Done Along the x-axis

The first part of the motion is along the x-axis from \((0, 0)\) to \((a, 0)\). The force vector when \(y=0\) is \(-K(0 \hat{\mathbf{i}} + x \hat{\mathbf{j}}) = -Kx \hat{\mathbf{j}}\). Since the motion is parallel to the x-axis, \(d\mathbf{r} = dx \hat{\mathbf{i}}\). Therefore, the work done \(W_x\) is the integral of \(\mathbf{F} \cdot d\mathbf{r}\) along this path:\[W_x = \int_0^a (-Kx \hat{\mathbf{j}}) \cdot (dx \hat{\mathbf{i}}) = 0.\]Since the dot product between \(\hat{\mathbf{j}}\) and \(\hat{\mathbf{i}}\) is zero, no work is done along the x-axis.

Step 2: Determine Work Done Along the y-axis

The second part of the motion is along the y-axis from \((a, 0)\) to \((a, a)\). The force vector when \(x=a\) is \(-K(y \hat{\mathbf{i}} + a \hat{\mathbf{j}})\). The motion is parallel to the y-axis, so \(d\mathbf{r} = dy \hat{\mathbf{j}}\). Therefore, the work done \(W_y\) is:\[W_y = \int_0^a (-K(y \hat{\mathbf{i}} + a \hat{\mathbf{j}})) \cdot (dy \hat{\mathbf{j}}) = \int_0^a (-Ka) \, dy = -Ka \cdot a = -Ka^2.\]

Step 3: Calculate Total Work Done

The total work done \(W_{total}\) is the sum of the work done along the x-axis \(W_x\) and the y-axis \(W_y\):\[W_{total} = W_x + W_y = 0 - Ka^2 = -Ka^2.\]Thus, the total work done by the force on the particle is \(-Ka^2\).

Key Concepts

Force Vector

In physics, a force vector is an expression that not only has a magnitude (which is how much force is being applied) but also a direction. In scenarios where forces are represented as vectors, they can be broken down into components along specified axes, like the x-axis and y-axis. This is useful because it allows us to understand how a force affects an object in a multidimensional space. For example, consider a force given by \( \mathbf{F} = -K(y \hat{\mathbf{i}} + x \hat{\mathbf{j}}) \). This force vector is outlined in terms of its components along the x-direction (\( \hat{\mathbf{i}} \)) and y-direction (\( \hat{\mathbf{j}} \)).

• The \( y \hat{\mathbf{i}} \) component influences movement or effects along the x-axis.
• Similarly, the \( x \hat{\mathbf{j}} \) component affects the y-axis.

Understanding force vectors is crucial for analyzing physical scenarios, as it helps in determining how different components of forces act on a particle as it moves. Force vectors also help in visualizing action, for example in 2D planes such as the x and y plane utilized in this problem.

Work-Energy Theorem

The work-energy theorem is an important concept that connects the amount of work done on an object to its change in kinetic energy. The theorem essentially states that the work done by all forces acting on a particle equals the change in the particle's kinetic energy. This relationship can be expressed mathematically as:\[W_{total} = \Delta KE = KE_{final} - KE_{initial},\]where \( W_{total} \) is the total work done by all forces. However, in the given exercise, we are focused on calculating the work done by a specific force as the particle follows a path in the xy-plane, not directly analyzing energy change.

In this specific example, calculating separate work contributions along the different segments of the path and then summing them gives the total work done by the force:\[W_{total} = W_x + W_y.\]

The theorem aids in conceptualizing how forces do work during the movement of objects along predefined pathways, but here, in the step-by-step completion, the focus was on understanding how each segment's work contributes to the overall work.

Integral Calculus in Physics

Integral calculus plays a significant role in physics, especially when dealing with continuous processes or cumulative sums over paths. When a force varies along a path, calculating work requires evaluating an integral. The general expression for work done by a force \(\mathbf{F}\) along a path \(\mathbf{d}\mathbf{r}\) is given by the integral of the dot product:\[W = \int_{path} \mathbf{F} \cdot d\mathbf{r}.\]

In this exercise, the force vector changes as the particle follows its path, requiring to break down paths and solve each integral separately. For example:

• Along the x-axis (from \((0, 0)\) to \((a, 0)\)), the dot product results in zero work due to perpendicular components (\( \hat{\mathbf{i}} \cdot \hat{\mathbf{j}} = 0 \)).
• Along the y-axis (from \((a, 0)\) to \((a, a)\)), the integration results in meaningful work done, calculated as the product \(-Ka^2\).

This usage of integral calculus helps us make precise calculations about how forces act over curvy or complex movement paths.

Path of Motion in xy-plane

When considering motion in physics, especially in 2D like the xy-plane, it is essential to appreciate how paths influence interactions of forces and particles. In the given problem, the particle moves first along the x-axis and then along the y-axis. Understanding this path helps in assessing how the force components affect the particle, as each axis can be treated independently under typical conditions.

• In the first segment (x-axis), since motion is solely along the x-axis, the component of force parallel to this motion is negligible, resulting in zero work done.
• In the second segment (y-axis), the motion is affected by the component \(-Kx \hat{\mathbf{j}}\), leading to the non-zero work calculated by integrating across the path.

Tracking the paths in such problems is critical, as the path directly influences how different forces apply, demonstrating how line integrals are used to evaluate work where direct application at a point isn't plainly sufficient.