Concept Card: Work-Energy Theorem for Constant & Variable Forces
1. Statement & Universal Scope of the Work-Energy Theorem (WET):
The net work done by all external and internal forces acting on a body (or system of bodies) equals the total change in its kinetic energy:
$$\mathbf{W_{\text{net}} = \Delta K = K_f – K_i = \frac{1}{2}m v_f^2 – \frac{1}{2}m v_i^2}$$
The “All Forces” Principle:
Unlike the Law of Conservation of Mechanical Energy (which strictly holds only when non-conservative forces do zero work), the Work-Energy Theorem is universally valid without exception:
$$W_{\text{conservative}} + W_{\text{non-conservative}} + W_{\text{pseudo}} + W_{\text{external}} = \Delta K$$
Every force in nature—conservative (gravity, spring, electrostatic), non-conservative (friction, drag), external, internal, and pseudo forces in accelerating frames—contributes to the total work.
2. Mathematical Derivations:
- Constant Net Force (1D Motion):
From kinematics: $v_f^2 = v_i^2 + 2ad \implies ad = \frac{v_f^2 – v_i^2}{2}$.
Multiplying by mass $m$:
$F_{\text{net}} d = m(ad) = m\left(\frac{v_f^2 – v_i^2}{2}\right) = \frac{1}{2}mv_f^2 – \frac{1}{2}mv_i^2 \implies W_{\text{net}} = \Delta K$. - Constant Net Force (3D Vector Motion):
$W_{\text{net}} = \vec{F}_{\text{net}} \cdot \Delta\vec{r} = m\left(\frac{\vec{v}_f – \vec{v}_i}{t}\right) \cdot \left(\frac{\vec{v}_i + \vec{v}_f}{2}\right)t = \frac{1}{2}m(v_f^2 – v_i^2) = \Delta K$. - Variable Force (Calculus Proof):
$W = \int_{C} \vec{F} \cdot d\vec{r} = \int_{t_i}^{t_f} \left(m\frac{d\vec{v}}{dt}\right) \cdot (\vec{v}\,dt) = \int_{t_i}^{t_f} \frac{d}{dt}\left(\frac{1}{2}m v^2\right)dt = \left[\frac{1}{2}mv^2\right]_{v_i}^{v_f} = \Delta K$.
3. Work-Energy Theorem in Non-Inertial Reference Frames:
In a reference frame translating with linear acceleration $\vec{a}_0$:
$$W_{\text{real}} + W_{\text{pseudo}} = \Delta K_{\text{rel}}$$
where $W_{\text{pseudo}} = \int (-m\vec{a}_0) \cdot d\vec{r}_{\text{rel}}$.
Axiom: Both work and kinetic energy are frame-dependent, but WET holds consistently in every reference frame.
4. Standard Calculation Checklist for Work Done by Common Forces:
- Gravity: $W_g = -mg\Delta y = mg(y_i – y_f)$ (independent of path followed).
- Normal Force: $W_N = 0$ (whenever displacement is parallel to contact plane).
- Friction: $W_f = -\int f_k\,ds = -\mu_k mg\cos\theta \cdot s = -\mu_k mg \cdot x_{\text{horizontal}}$ for any flat or inclined profile.
- Ideal Spring: $W_s = -\frac{1}{2}k(x_f^2 – x_i^2)$.
Solved Examples
Example 1 (Stopping Distance & Bullet Penetration into Planks):
A bullet of mass $m = 10.0\text{ g}$ moving horizontally with initial speed $v_0 = 400.0\text{ m/s}$ strikes a wooden block and penetrates $d = 20.0\text{ cm}$ ($0.20\text{ m}$) before coming to rest.
(a) Determine the average resistive force exerted by the wood.
(b) If an identical bullet strikes a wooden plank of thickness $t = 15.0\text{ cm}$ with the same initial speed, find its exit speed.
(c) How many such planks of thickness $15.0\text{ cm}$ are required to stop the bullet completely?
Solution:
(a) Initial kinetic energy: $K_i = \frac{1}{2}(0.010)(400.0^2) = 800.0\text{ Joules}$. Final kinetic energy $K_f = 0$.
By WET: $-F_{\text{avg}} d = 0 – 800.0 \implies F_{\text{avg}}(0.20) = 800.0 \implies F_{\text{avg}} = 4000.0\text{ N} = 4.0\text{ kN}$.
(b) For thickness $t = 0.15\text{ m}$: $W_{\text{plank}} = -F_{\text{avg}} t = -4000.0 \times 0.15 = -600.0\text{ Joules}$.
By WET: $-600.0 = \frac{1}{2}(0.010)v_1^2 – 800.0 \implies 0.005 v_1^2 = 200.0 \implies v_1^2 = 40000 \implies v_1 = 200.0\text{ m/s}$.
(c) Total stopping distance is $20.0\text{ cm}$. Number of $15.0\text{ cm}$ planks needed: $N = \frac{20.0}{15.0} = 1.33 \implies 2\text{ planks}$.
Example 2 (Rough Incline with Spring Buffer):
A block of mass $m = 2.0\text{ kg}$ is released from rest at the top of a rough incline of angle $\theta = 30^\circ$. The distance along the incline to an unstretched spring of stiffness $k = 400\text{ N/m}$ is $L = 2.0\text{ m}$. The coefficient of kinetic friction is $\mu_k = \frac{0.20}{\sqrt{3}}$. Taking $g = 10\text{ m/s}^2$:
(a) Formulate the WET equation from release to maximum spring compression $x_{\text{max}}$.
(b) Calculate $x_{\text{max}}$.
Solution:
(a) Total displacement along incline is $s = L + x_{\text{max}}$.
$W_g = mg(L + x_{\text{max}})\sin(30^\circ) = 2.0(10)(L + x_{\text{max}})(0.5) = 10(L + x_{\text{max}})$.
$W_f = -(\mu_k mg\cos 30^\circ)(L + x_{\text{max}}) = -\left(\frac{0.20}{\sqrt{3}}\right)(20)\left(\frac{\sqrt{3}}{2}\right)(L + x_{\text{max}}) = -2.0(L + x_{\text{max}})$.
$W_s = -\frac{1}{2}k x_{\text{max}}^2 = -200 x_{\text{max}}^2$.
Since $v_i = 0$ and $v_f = 0 \implies \Delta K = 0$:
$W_g + W_f + W_s = 0 \implies 8.0(L + x_{\text{max}}) – 200 x_{\text{max}}^2 = 0$.
(b) With $L = 2.0\text{ m}$: $16.0 + 8.0 x_{\text{max}} = 200 x_{\text{max}}^2 \implies 25 x_{\text{max}}^2 – x_{\text{max}} – 2.0 = 0$.
$x_{\text{max}} = \frac{1 + \sqrt{1 + 200}}{50} = \frac{1 + \sqrt{201}}{50} \approx \frac{1 + 14.177}{50} = 0.3035\text{ m} \approx 30.4\text{ cm}$.
Example 3 (Variable Force on a Vertical Circular Track):
A bead of mass $m = 0.50\text{ kg}$ slides on a smooth vertical circular hoop of radius $R = 1.0\text{ m}$, starting from rest at the top $A$. As it slides to bottom $B$, a constant tangential force $F_t = 5.0\text{ N}$ drives it forward along the arc. Taking $g = 10\text{ m/s}^2$:
(a) Calculate the work done by gravity, normal force, and tangential force.
(b) Find the speed $v_B$ at bottom $B$.
(c) Find the normal force on the bead at $B$.
Solution:
(a) $W_g = mg(2R) = 0.50(10)(2.0) = 10.0\text{ Joules}$.
$W_N = 0$ (normal force is radial, $\vec{N} \perp d\vec{s}$).
$W_{F_t} = F_t(\pi R) = 5.0(\pi \times 1.0) = 5\pi \approx 15.71\text{ Joules}$.
$W_{\text{net}} = 10.0 + 5\pi \approx 25.71\text{ Joules}$.
(b) By WET: $25.71 = \frac{1}{2}(0.50)v_B^2 = 0.25 v_B^2 \implies v_B = \sqrt{102.83} \approx 10.14\text{ m/s}$.
(c) Normal force at the bottom: $N – mg = \frac{m v_B^2}{R} \implies N = 0.50(10) + \frac{0.50(102.83)}{1.0} = 5.0 + 51.42 = 56.42\text{ N}$.
Example 4 (Work-Energy Theorem in an Accelerating Elevator Frame):
A smooth wedge of inclination $\theta = 30^\circ$ is inside an elevator accelerating upward with $a_0 = 2.0\text{ m/s}^2$. A block of mass $m = 2.0\text{ kg}$ is launched up the incline with speed $u = 6.0\text{ m/s}$ relative to the wedge. Taking $g = 10\text{ m/s}^2$:
(a) Using WET in the elevator frame, find the maximum distance $s$ along the incline.
(b) Verify the result in the ground frame.
Solution:
(a) In elevator frame, downward pseudo-force is $m a_0$. Effective gravity: $g_{\text{eff}} = g + a_0 = 12.0\text{ m/s}^2$.
$W_{\text{net}} = -m(g + a_0)s\sin(30^\circ) = -2.0(12.0)s(0.50) = -12.0 s\text{ Joules}$.
By WET: $W_{\text{net}} = 0 – \frac{1}{2}m u^2 = -\frac{1}{2}(2.0)(6.0^2) = -36.0\text{ Joules}$.
$-12.0 s = -36.0 \implies s = 3.0\text{ meters}$.
(b) Ground frame check: Deceleration along the plane is $a_{\text{rel}} = (g + a_0)\sin(30^\circ) = 6.0\text{ m/s}^2$.
$s = \frac{u^2}{2a_{\text{rel}}} = \frac{36.0}{2(6.0)} = 3.0\text{ meters}$.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
A bullet fired into a fixed wooden target loses $\frac{1}{n}$ of its velocity in penetrating distance $x$. Assuming uniform resistance, how much further will it penetrate before coming to rest?
(A) $\frac{x}{n^2 – 1}$
(B) $\frac{(n – 1)^2 x}{2n – 1}$
(C) $\frac{x}{2n – 1}$
(D) $\frac{n^2 x}{n – 1}$
Problem 2 (JEE Main – Single Correct):
A block of mass $m = 2.0\text{ kg}$ is released from rest at the top of a rough curved track of height $h = 5.0\text{ m}$. It arrives at the bottom with speed $v = 8.0\text{ m/s}$. Taking $g = 10\text{ m/s}^2$, the work done by friction is:
(A) $-100\text{ J}$
(B) $-64\text{ J}$
(C) $-36\text{ J}$
(D) $-50\text{ J}$
Problem 3 (JEE Main – Single Correct):
A body of mass $m$ initially at rest is acted upon by a 1D force $F(x) = \alpha x$. The velocity after moving through distance $d$ is:
(A) $\sqrt{\frac{\alpha d^2}{m}}$
(B) $\sqrt{\frac{2\alpha d^2}{m}}$
(C) $\frac{\alpha d}{m}$
(D) $\sqrt{\frac{\alpha d}{2m}}$
Problem 4 (JEE Main – Single Correct):
A car traveling at speed $v$ is stopped in a minimum braking distance $d$. If the same car travels at speed $3v$, the minimum stopping distance under the same retarding force is:
(A) $3d$
(B) $6d$
(C) $9d$
(D) $\sqrt{3}d$
Problem 5 (JEE Main – Single Correct):
A simple pendulum of length $L$ and bob mass $m$ is released from a horizontal orientation ($\theta = 90^\circ$ to vertical). By the Work-Energy Theorem, the speed of the bob at the lowest point is:
(A) $\sqrt{g L}$
(B) $\sqrt{2 g L}$
(C) $2\sqrt{g L}$
(D) $\sqrt{\frac{g L}{2}}$
Problem 6 (JEE Advanced – One or More Correct):
Which of the following statements is/are correct regarding the Work-Energy Theorem (WET)?
(A) The Work-Energy Theorem is valid for conservative, non-conservative, internal, and external forces alike.
(B) In a non-inertial reference frame, the Work-Energy Theorem remains valid provided the work done by pseudo forces is included.
(C) The net work done on a particle is zero if and only if its speed is constant.
(D) Since kinetic energy depends on the chosen reference frame, the work done by a force can be frame-dependent.
Problem 7 (JEE Advanced – One or More Correct):
A block of mass $m$ is pushed against a horizontal spring of stiffness $k$, compressing it by $x_0$. When released from rest on a rough horizontal floor with kinetic friction coefficient $\mu$:
(A) The work done by the spring force when the block reaches the uncompressed position is $+\frac{1}{2}k x_0^2$.
(B) The work done by friction as the block moves distance $x_0$ to the natural length position is $-\mu mg x_0$.
(C) The speed of the block at the natural length position is $\sqrt{\frac{k}{m}x_0^2 – 2\mu g x_0}$.
(D) The block will oscillate indefinitely about the equilibrium position.
Problem 8 (JEE Advanced – One or More Correct):
A particle of mass $m$ moves along the $x$-axis from $x = 0$ to $x = 4\text{ m}$ under a force $F(x) = (2 – x)\text{ N}$. If the particle was initially at rest at $x = 0$:
(A) The kinetic energy of the particle is maximum at $x = 2.0\text{ m}$.
(B) The maximum kinetic energy attained is $2.0\text{ J}$.
(C) The particle comes to instantaneous rest at $x = 4.0\text{ m}$.
(D) The net work done by the force from $x = 0$ to $x = 4.0\text{ m}$ is zero.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A block of mass $m = 1.0\text{ kg}$ moving with an initial speed $v_0 = 10.0\text{ m/s}$ enters a rough horizontal patch of length $L = 8.0\text{ m}$ where $\mu_k = 0.40$. Taking $g = 10\text{ m/s}^2$, find the speed of the block in $\text{m/s}$ as it leaves the rough patch.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A bullet of mass $m = 0.020\text{ kg}$ traveling at $v = 200.0\text{ m/s}$ strikes a sandbag and penetrates a distance $d = 0.40\text{ m}$ before stopping. The magnitude of the average retarding force in kiloNewtons ($\text{kN}$) is $F$. Find $F$.
Solutions & Explanations
Answer Key Summary:
1. (B) | 2. (C) | 3. (A) | 4. (C) | 5. (B) | 6. (A, B, C, D) | 7. (A, B, C) | 8. (A, B, C, D) | 9. 6 | 10. 1
Solution 1:
$v_1 = v_0\left(\frac{n-1}{n}\right)$. In first stage: $-Fx = \frac{1}{2}m v_0^2\left[\left(\frac{n-1}{n}\right)^2 – 1\right] = -\frac{1}{2}mv_0^2\left(\frac{2n-1}{n^2}\right)$. For further distance $y$: $-Fy = -\frac{1}{2}mv_1^2 = -\frac{1}{2}mv_0^2\left(\frac{n-1}{n}\right)^2$. Dividing gives $y = \frac{(n-1)^2 x}{2n-1}$.
Correct Answer: (B)
Solution 2:
$W_g + W_f = K_f – K_i \implies 2.0(10)(5.0) + W_f = \frac{1}{2}(2.0)(8.0^2) \implies 100.0 + W_f = 64.0 \implies W_f = -36.0\text{ J}$.
Correct Answer: (C)
Solution 3:
$W = \int_0^d \alpha x\,dx = \frac{1}{2}\alpha d^2 = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{\alpha d^2}{m}}$.
Correct Answer: (A)
Solution 4:
$d \propto v^2$. Tripling speed increases stopping distance by $3^2 = 9$ times.
Correct Answer: (C)
Solution 5:
$W_T = 0$ (orthogonal to motion). $W_g = mgL = \frac{1}{2}mv^2 \implies v = \sqrt{2gL}$.
Correct Answer: (B)
Solution 6:
All statements (A, B, C, D) are verified foundational principles of the Work-Energy Theorem.
Correct Answer: (A, B, C, D)
Solution 7:
(A) $W_s = \frac{1}{2}kx_0^2$ (True). (B) $W_f = -\mu mg x_0$ (True). (C) By WET: $\frac{1}{2}mv^2 = \frac{1}{2}kx_0^2 – \mu mg x_0 \implies v = \sqrt{\frac{k}{m}x_0^2 – 2\mu gx_0}$ (True). (D) Friction dissipates mechanical energy, damping oscillations to rest (False).
Correct Answer: (A, B, C)
Solution 8:
All statements (A, B, C, D) describe the complete calculus and energy properties of $F(x) = 2 – x$.
Correct Answer: (A, B, C, D)
Solution 9:
$v_f^2 = v_0^2 – 2\mu_k g L = 100.0 – 2(0.40)(10)(8.0) = 100.0 – 64.0 = 36.0 \implies v_f = 6.0\text{ m/s}$.
Correct Answer: 6
Solution 10:
$-F d = -\frac{1}{2}mv^2 \implies F(0.40) = \frac{1}{2}(0.020)(40000) = 400.0\text{ J} \implies F = 1000\text{ N} = 1.0\text{ kN}$.
Correct Answer: 1