Tag: Work done by a variable force

Summary:

This is lecture 2 for the topic Work, Energy and Power. You can find notes of lecture 1 for this topic here.

1. Why $W = \vec{F} \cdot \vec{s}$ Fails

The formula $W = \vec{F} \cdot \vec{s}$ only applies when the force is strictly constant in both magnitude and direction over the entire displacement. However, in the real world (and in JEE problems), forces frequently change as an object moves:

• Springs: The force increases as you stretch it ($F = -kx$).
• Gravity (Large distances): The force weakens as you move away from a planet ($F = \frac{GMm}{r^2}$).
• Time/Position-dependent fields: Electric or magnetic fields that vary across space.

If the force varies, we cannot use a single value for $\vec{F}$. We must break the displacement into infinitesimally small steps where the force can be considered constant.

2. The Integration Method (Calculus Approach)

Assume a particle undergoes an infinitesimally small displacement $d\vec{r}$. Over this tiny distance, a variable force $\vec{F}$ is effectively constant.

The infinitesimally small work done $dW$ is:

$$dW = \vec{F} \cdot d\vec{r}$$

To find the total work done from an initial position $\vec{r}_i$ to a final position $\vec{r}_f$, we sum up all these tiny contributions using integration:

$$W = \int_{\vec{r}_i}^{\vec{r}_f} \vec{F} \cdot d\vec{r}$$

Cartesian (3D) Form:

In JEE problems, $\vec{F}$ is often given as a function of coordinates: $\vec{F} = F_x \hat{i} + F_y \hat{j} + F_z \hat{k}$.

The infinitesimal displacement is $d\vec{r} = dx \hat{i} + dy \hat{j} + dz \hat{k}$.

The dot product expands the integral into three separate 1D integrals:

$$W = \int_{x_i}^{x_f} F_x dx + \int_{y_i}^{y_f} F_y dy + \int_{z_i}^{z_f} F_z dz$$

(Note: For conservative forces, this integration is path-independent. For non-conservative forces like friction, it depends on the specific path taken.)

3. The Graphical Method (Area Under the Curve)

Geometrically, the definite integral $\int F dx$ represents the area bounded by the Force vs. Position curve and the position axis.

Key Rules for Graphical Problems:

1. Work = Area under the $F-x$ graph.
2. Sign Convention:

• Area above the x-axis ($F > 0$) represents positive work (force and displacement are in the same direction; kinetic energy increases).
• Area below the x-axis ($F < 0$) represents negative work (force opposes displacement; kinetic energy decreases).

3. Total Work: $W_{total} = (\text{Area above axis}) – (\text{Area below axis})$.

4. Classic Application: Work Done by a Spring

A perfectly elastic spring obeys Hooke’s Law: $F_s = -kx$.

If we stretch a spring from its natural length ($x_i = 0$) to a position $x_f = x$, the work done by the spring force is:

$$W_s = \int_{0}^{x} (-kx) dx = -k \left[ \frac{x^2}{2} \right]_0^x = -\frac{1}{2}kx^2$$

The negative sign indicates the spring force always opposes the displacement (stretching or compressing).

Conversely, the work done by the external applied force (which must be $+kx$ to stretch it slowly) is:

$$W_{ext} = \int_{0}^{x} (kx) dx = +\frac{1}{2}kx^2$$

JEE-Mains Practice Questions

Question 1: 1D Integration

A particle of mass $2 \text{ kg}$ is subjected to a one-dimensional variable force $F(x) = (3x^2 – 2x + 1) \text{ N}$. Calculate the work done by this force in moving the particle from $x = 1 \text{ m}$ to $x = 3 \text{ m}$.

Solution:

$W = \int_{x_i}^{x_f} F(x) dx$

$W = \int_{1}^{3} (3x^2 – 2x + 1) dx$

$W = \left[ x^3 – x^2 + x \right]_1^3$

Substitute upper limit: $(3^3 – 3^2 + 3) = 27 – 9 + 3 = 21$

Substitute lower limit: $(1^3 – 1^2 + 1) = 1 – 1 + 1 = 1$

$W = 21 – 1 = 20 \text{ J}$

Question 2: 2D Force Field

A force $\vec{F} = (y\hat{i} + x\hat{j}) \text{ N}$ acts on a particle. Find the work done by this force in moving the particle from the origin $(0,0)$ to the point $(2 \text{ m}, 2 \text{ m})$ along the straight line path $y = x$.

Solution:

Since the path is $y = x$, it follows that $dy = dx$.

We can substitute $y$ with $x$ in our integral to make it a single variable integration.

$W = \int (F_x dx + F_y dy)$

$W = \int (y dx + x dy)$

Substitute $y=x$ and $dy=dx$:

$W = \int_{0}^{2} (x dx + x dx) = \int_{0}^{2} 2x dx$

$W = \left[ x^2 \right]_0^2 = 2^2 – 0^2 = 4 \text{ J}$

(Note: This force is conservative, $\vec{F} = \nabla(xy)$, so $W$ is just $\Delta(xy) = (2)(2) – (0)(0) = 4 \text{ J}$. Teaching both methods is highly recommended!)

Question 3: Graphical Method

The force acting on a particle varies with position $x$ as shown in a graph. The graph consists of a straight line from $(0\text{ m}, 10\text{ N})$ to $(4\text{ m}, 10\text{ N})$, followed by a straight line dropping to $(8\text{ m}, 0\text{ N})$, and then continuing to $(10\text{ m}, -5\text{ N})$. What is the total work done from $x = 0$ to $x = 10 \text{ m}$?

Solution:

We calculate the area of three geometric shapes:

1. Rectangle from $x=0$ to $4$: Area $= 4 \times 10 = 40 \text{ J}$
2. Triangle from $x=4$ to $8$: Area $= \frac{1}{2} \times (8-4) \times 10 = \frac{1}{2} \times 4 \times 10 = 20 \text{ J}$
3. Triangle from $x=8$ to $10$ (below axis): Area $= \frac{1}{2} \times (10-8) \times (-5) = \frac{1}{2} \times 2 \times (-5) = -5 \text{ J}$

$W_{total} = 40 + 20 + (-5) = 55 \text{ J}$