Concept Card: Work Done by Constant & Variable Forces & Kinetic Energy
1. Physical Definition of Mechanical Work:
Work is done by a force when its point of application undergoes a displacement having a non-zero component along the line of action of the force.
Work Done by a Constant Force ($\vec{F}$):
Defined as the scalar (dot) product of the constant force vector $\vec{F}$ and the displacement vector $\vec{d}$:
$$W = \vec{F} \cdot \vec{d} = F d \cos\theta$$
where $\theta$ is the angle between $\vec{F}$ and $\vec{d}$.
In Cartesian component form: $W = F_x \Delta x + F_y \Delta y + F_z \Delta z$.
– SI Unit: Joule ($\text{J} = \text{N}\cdot\text{m} = \text{kg}\cdot\text{m}^2/\text{s}^2$).
– CGS Unit: $\text{erg} = 10^{-7}\text{ J}$. Dimensions: $[M L^2 T^{-2}]$.
2. Signs and Classifications of Work:
- Positive Work ($0 \le \theta < 90^\circ$): Force component aligns with displacement ($\cos\theta > 0$); mechanical energy is delivered to the body (e.g., gravity on a falling mass). Maximum at $\theta = 0^\circ$ ($W = F d$).
- Zero Work ($W = 0$): Occurs when:
- Displacement is zero ($d = 0$, e.g., pushing against a rigid wall).
- Force is zero ($F = 0$).
- Force and displacement are mutually orthogonal ($\theta = 90^\circ$, $\vec{F} \perp \vec{d}$):
- Normal contact reaction on a body sliding on a horizontal plane.
- Tension in a string during simple or conical pendulum motion.
- Centripetal force in uniform circular motion ($W_c = 0$).
- Magnetic Lorentz force on a moving charge: $\vec{F}_B = q(\vec{v}\times\vec{B}) \perp \vec{v} \implies W_B = 0$.
- Negative Work ($90^\circ < \theta \le 180^\circ$): Force component opposes displacement ($\cos\theta < 0$); mechanical energy is removed from the body (e.g., kinetic friction, air resistance). Maximum negative work occurs at $\theta = 180^\circ$ ($W = -F d$).
3. Work Done by a Variable Force:
When the force depends on position $\vec{F}(\vec{r})$:
$$W = \int_{\vec{r}_i}^{\vec{r}_f} \vec{F} \cdot d\vec{r} = \int_{x_i}^{x_f} F_x\,dx + \int_{y_i}^{y_f} F_y\,dy + \int_{z_i}^{z_f} F_z\,dz$$
- Graphical Area Method ($F-x$ Graph): For one-dimensional motion, work equals the area under the $F(x)$ curve bounded by the initial and final positions:
$W = \int_{x_i}^{x_f} F(x)\,dx = \text{Area under } F-x \text{ curve}$
Areas above the displacement axis are positive; areas below are negative. - Work Done by an Ideal Spring: For restoring force $F_s = -kx$:
$W_{\text{spring}} = \int_{x_1}^{x_2} (-kx)\,dx = -\frac{1}{2}k(x_2^2 – x_1^2)$
Stretching from rest ($0 \to x$): $W_{\text{spring}} = -\frac{1}{2}kx^2$, while the external pulling agent does $W_{\text{ext}} = +\frac{1}{2}kx^2$.
4. Kinetic Energy ($K$):
Kinetic energy is the mechanical energy possessed by an object due to its translational motion:
$$K = \frac{1}{2}m v^2 = \frac{1}{2}m (\vec{v}\cdot\vec{v})$$
- Strictly a non-negative scalar quantity ($K \ge 0$).
- Frame-Dependent: Because velocity $\vec{v}$ depends on the chosen reference frame, kinetic energy is relative to the observer’s frame.
- Relation with Linear Momentum:
$K = \frac{p^2}{2m} \iff p = \sqrt{2mK}$
For small variations ($\le 5\%$): $\frac{\Delta K}{K} \approx 2\frac{\Delta p}{p}$.
Solved Examples
Example 1 (Work Done by a Constant 3D Force Vector):
A constant force $\vec{F} = (4.0\hat{i} – 3.0\hat{j} + 2.0\hat{k})\text{ N}$ acts on a particle of mass $m = 2.0\text{ kg}$, displacing it from an initial position $\vec{r}_1 = (2.0\hat{i} + 3.0\hat{j} – 1.0\hat{k})\text{ m}$ to a final position $\vec{r}_2 = (6.0\hat{i} – 1.0\hat{j} + 3.0\hat{k})\text{ m}$.
(a) Determine the displacement vector $\Delta\vec{r}$.
(b) Calculate the work done by the force $\vec{F}$.
(c) Find the angle $\theta$ between the force vector and the displacement vector.
Solution:
(a) $\Delta\vec{r} = \vec{r}_2 – \vec{r}_1 = (6.0 – 2.0)\hat{i} + (-1.0 – 3.0)\hat{j} + (3.0 – (-1.0))\hat{k} = 4.0\hat{i} – 4.0\hat{j} + 4.0\hat{k}\text{ m}$.
(b) $W = \vec{F} \cdot \Delta\vec{r} = (4.0)(4.0) + (-3.0)(-4.0) + (2.0)(4.0) = 16.0 + 12.0 + 8.0 = 36.0\text{ Joules}$.
(c) $|\vec{F}| = \sqrt{4.0^2 + (-3.0)^2 + 2.0^2} = \sqrt{29}\text{ N} \approx 5.385\text{ N}$.
$|\Delta\vec{r}| = \sqrt{4.0^2 + (-4.0)^2 + 4.0^2} = \sqrt{48} = 4\sqrt{3}\text{ m} \approx 6.928\text{ m}$.
$\cos\theta = \frac{W}{|\vec{F}||\Delta\vec{r}|} = \frac{36.0}{\sqrt{29} \times 4\sqrt{3}} = \frac{9.0}{\sqrt{87}} \approx 0.9649 \implies \theta \approx 15.22^\circ$.
Example 2 (Calculus Integration & Geometric F-x Area Evaluation):
(a) A 1D variable force $F(x) = (3x^2 – 4x + 5)\text{ N}$ acts on a particle moving along the $x$-axis from $x = 0$ to $x = 4.0\text{ m}$. Calculate the work done.
(b) A body is subjected to a piecewise force $F(x)$: increases linearly from $0$ to $30.0\text{ N}$ as $x$ goes from $0$ to $2.0\text{ m}$; remains constant at $30.0\text{ N}$ from $x = 2.0\text{ m}$ to $5.0\text{ m}$; drops linearly to $-10.0\text{ N}$ at $x = 7.0\text{ m}$ (crossing $F = 0$ at $x = 6.5\text{ m}$). Find the net work done from $x = 0$ to $x = 7.0\text{ m}$.
Solution:
(a) $W = \int_{0}^{4} (3x^2 – 4x + 5)\,dx = \left[x^3 – 2x^2 + 5x\right]_{0}^{4} = (64 – 32 + 20) = 52.0\text{ Joules}$.
(b) Partition into geometric areas:
– $A_1$ (Triangle $0 \to 2.0\text{ m}$): $\frac{1}{2}(2.0)(30.0) = 30.0\text{ J}$.
– $A_2$ (Rectangle $2.0 \to 5.0\text{ m}$): $(3.0)(30.0) = 90.0\text{ J}$.
– $A_3$ (Positive triangle $5.0 \to 6.5\text{ m}$): $\frac{1}{2}(1.5)(30.0) = 22.5\text{ J}$.
– $A_4$ (Negative triangle $6.5 \to 7.0\text{ m}$): $\frac{1}{2}(0.5)(-10.0) = -2.5\text{ J}$.
$W_{\text{net}} = 30.0 + 90.0 + 22.5 – 2.5 = 140.0\text{ Joules}$.
Example 3 (Spring Compression & Kinetic Energy Conversion):
A block of mass $m = 1.0\text{ kg}$ sliding on a smooth horizontal floor with speed $v_0 = 4.0\text{ m/s}$ impacts a horizontal spring of stiffness $k = 400\text{ N/m}$ fixed to a wall.
(a) Express work done by the spring force as a function of compression $x$.
(b) Calculate the maximum compression $x_{\text{max}}$ of the spring.
(c) Find the speed of the block when the spring is compressed by $x = 10.0\text{ cm}$.
Solution:
(a) $W_s(x) = \int_{0}^{x} (-kx’)\,dx’ = -\frac{1}{2}kx^2$.
(b) At maximum compression, $v = 0$. Initial kinetic energy is fully converted into spring potential energy:
$\frac{1}{2}m v_0^2 = \frac{1}{2}k x_{\text{max}}^2 \implies x_{\text{max}} = v_0\sqrt{\frac{m}{k}} = 4.0\sqrt{\frac{1.0}{400}} = 4.0\left(\frac{1}{20}\right) = 0.20\text{ m} = 20.0\text{ cm}$.
(c) At $x = 0.10\text{ m}$:
$\frac{1}{2}m v^2 = \frac{1}{2}m v_0^2 – \frac{1}{2}k x^2 \implies v^2 = v_0^2 – \frac{k}{m}x^2 = 16.0 – 400(0.01) = 12.0\text{ m}^2/\text{s}^2$.
$v = \sqrt{12.0} = 2\sqrt{3}\text{ m/s} \approx 3.464\text{ m/s}$.
Example 4 (Path Independence & 2D Conservative Force Fields):
A particle moves in the $x-y$ plane under the force $\vec{F} = (2xy)\hat{i} + (x^2)\hat{j}\text{ N}$.
(a) Prove that the force field is conservative and find its potential function $U(x, y)$.
(b) Evaluate the work done moving from $(0, 0)$ to $(2, 3)$ along Path 1 ($y = \frac{3}{4}x^2$) and Path 2 ($y = \frac{3}{2}x$).
(c) Evaluate the work done around any closed loop.
Solution:
(a) $\frac{\partial F_y}{\partial x} = \frac{\partial}{\partial x}(x^2) = 2x, \quad \frac{\partial F_x}{\partial y} = \frac{\partial}{\partial y}(2xy) = 2x \implies \vec{\nabla}\times\vec{F} = \vec{0}$. The force is conservative.
Potential function: $\vec{F} = -\vec{\nabla}U \implies U(x, y) = -x^2 y + C$.
(b) Along Path 1 ($dy = \frac{3}{2}x\,dx$):
$W = \int_{0}^{2} \left[2x\left(\frac{3}{4}x^2\right) + x^2\left(\frac{3}{2}x\right)\right]dx = \int_{0}^{2} 3x^3\,dx = \left[\frac{3x^4}{4}\right]_{0}^{2} = 12.0\text{ Joules}$.
Along Path 2 ($dy = \frac{3}{2}\,dx$):
$W = \int_{0}^{2} \left[2x\left(\frac{3}{2}x\right) + x^2\left(\frac{3}{2}\right)\right]dx = \int_{0}^{2} \frac{9}{2}x^2\,dx = \left[\frac{3}{2}x^3\right]_{0}^{2} = 12.0\text{ Joules}$.
Via potential difference: $W = -\Delta U = -[-(2^2)(3) – 0] = 12.0\text{ Joules}$.
(c) For any closed path in a conservative field: $\oint \vec{F} \cdot d\vec{r} = 0\text{ Joules}$.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
A particle of mass $m$ moves in a horizontal circle of radius $R$ with uniform speed $v$. The work done by the centripetal force during half a revolution is:
(A) $\pi R \left(\frac{m v^2}{R}\right)$
(B) $2 m v^2$
(C) $m v^2$
(D) Zero
Problem 2 (JEE Main – Single Correct):
If the kinetic energy of a moving particle is increased by $300\%$, the percentage increase in its linear momentum is:
(A) $50\%$
(B) $100\%$
(C) $150\%$
(D) $200\%$
Problem 3 (JEE Main – Single Correct):
A force $\vec{F} = (2\hat{i} + 3\hat{j})\text{ N}$ acts on a particle displacing it by $\vec{d} = (3\hat{i} + 4\hat{j})\text{ m}$. The work done by the force is:
(A) $6\text{ J}$
(B) $12\text{ J}$
(C) $18\text{ J}$
(D) $24\text{ J}$
Problem 4 (JEE Main – Single Correct):
A spring of natural length $L$ and spring constant $k$ is stretched from extension $x_1$ to extension $x_2$. The work done by the spring force is:
(A) $\frac{1}{2}k(x_2^2 – x_1^2)$
(B) $-\frac{1}{2}k(x_2^2 – x_1^2)$
(C) $\frac{1}{2}k(x_2 – x_1)^2$
(D) $-\frac{1}{2}k(x_2 – x_1)^2$
Problem 5 (JEE Main – Single Correct):
A satellite revolves around the Earth in an elliptical orbit. The total work done by the gravitational force on the satellite over one complete orbital period is:
(A) Zero
(B) Positive
(C) Negative
(D) Depends on orbital eccentricity
Problem 6 (JEE Advanced – One or More Correct):
Which of the following statements is/are correct regarding work and kinetic energy?
(A) Work is a scalar quantity that can be positive, negative, or zero.
(B) The kinetic energy of a body is strictly positive or zero in any reference frame.
(C) The numerical value of kinetic energy of a particle depends on the frame of reference.
(D) The normal contact force exerted by a surface can perform non-zero work on an object in an appropriate reference frame.
Problem 7 (JEE Advanced – One or More Correct):
For an ideal spring-mass system oscillating horizontally:
(A) The work done by the spring force over one complete cycle is zero.
(B) The spring force is a conservative force.
(C) While the spring is being compressed from its natural length, the work done by the spring force is negative.
(D) While the spring is expanding from maximum compression back to its natural length, the work done by the spring force is positive.
Problem 8 (JEE Advanced – One or More Correct):
A force acting on a particle moving along the $x$-axis is given by $F(x) = -kx + \beta x^3$ (where $k, \beta > 0$). Which of the following statements is/are correct?
(A) The work done by this force from $x = 0$ to $x = X$ is $-\frac{1}{2}k X^2 + \frac{1}{4}\beta X^4$.
(B) The force is conservative.
(C) The potential energy associated with this force (taking $U(0) = 0$) is $U(x) = \frac{1}{2}kx^2 – \frac{1}{4}\beta x^4$.
(D) The force is always directed towards the origin for all $x$.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A constant force $\vec{F} = (6.0\hat{i} + 8.0\hat{j})\text{ N}$ displaces a body from position $\vec{r}_1 = (1.0\hat{i} + 2.0\hat{j})\text{ m}$ to $\vec{r}_2 = (4.0\hat{i} + 0\hat{j})\text{ m}$. The work done by the force in Joules is $W$. Find $W$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A particle of mass $m = 2.0\text{ kg}$ initially at rest at $x = 0$ is subjected to a 1D position-dependent force $F(x) = 3.0x^2\text{ N}$. Find the kinetic energy of the particle in Joules when it reaches $x = 3.0\text{ m}$.
Solutions & Explanations
Answer Key Summary:
1. (D) | 2. (B) | 3. (C) | 4. (B) | 5. (A) | 6. (A, B, C, D) | 7. (A, B, C, D) | 8. (A, B, C) | 9. 2 | 10. 27
Solution 1:
Centripetal force is always orthogonal to displacement: $\vec{F}_c \cdot d\vec{s} = 0 \implies W = 0$.
Correct Answer: (D)
Solution 2:
$K_2 = K_1 + 3.00 K_1 = 4.00 K_1$. Since $p = \sqrt{2mK} \implies p_2 = \sqrt{2m(4K_1)} = 2p_1$. Percentage increase: $\frac{2p_1 – p_1}{p_1} \times 100\% = 100\%$.
Correct Answer: (B)
Solution 3:
$W = \vec{F} \cdot \vec{d} = (2)(3) + (3)(4) = 6 + 12 = 18\text{ J}$.
Correct Answer: (C)
Solution 4:
$W_{\text{spring}} = \int_{x_1}^{x_2} (-kx)\,dx = -\frac{1}{2}k(x_2^2 – x_1^2)$.
Correct Answer: (B)
Solution 5:
Gravitational force is conservative; work over any closed loop is identically zero.
Correct Answer: (A)
Solution 6:
All statements (A, B, C, D) are verified fundamental principles of work and kinetic energy (including positive normal work in an ascending elevator).
Correct Answer: (A, B, C, D)
Solution 7:
All statements (A, B, C, D) are verified properties of spring conservative mechanics.
Correct Answer: (A, B, C, D)
Solution 8:
– (A) $W = \int_{0}^{X} (-kx + \beta x^3)\,dx = -\frac{1}{2}kX^2 + \frac{1}{4}\beta X^4$ (True).
– (B) 1D force dependent solely on position is conservative (True).
– (C) $U(x) = -\int_{0}^{x} F(x’)\,dx’ = \frac{1}{2}kx^2 – \frac{1}{4}\beta x^4$ (True).
– (D) For $x > \sqrt{k/\beta}$, $\beta x^3 > kx \implies F > 0$, pointing away from origin (False).
Correct Answer: (A, B, C)
Solution 9:
$\Delta\vec{r} = (4.0 – 1.0)\hat{i} + (0 – 2.0)\hat{j} = 3.0\hat{i} – 2.0\hat{j}\text{ m}$.
$W = 6.0(3.0) + 8.0(-2.0) = 18.0 – 16.0 = 2.0\text{ Joules}$.
Correct Answer: 2
Solution 10:
$K = W = \int_{0}^{3.0} 3.0x^2\,dx = \left[x^3\right]_{0}^{3.0} = 27.0\text{ Joules}$.
Correct Answer: 27