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PhysicsLaws of Motion
Two cars A and B have kinetic energies of 100 J and 225 J respectively. Car A stops in 1000 m and car B stops in 1500 m after the brakes are applied. What is the ratio of the braking forces $F_A/F_B$?
Options
1
$3/2$
2
$2/3$
3
$1/1$
4
$4/9$
Correct Answer
$F_A/F_B = 2/3$
Solution
1

Work-energy theorem: $F \times d = KE$ (work done by brake = KE lost)

$$F_A = \frac{KE_A}{d_A} = \frac{100}{1000} = 0.1 \text{ N}$$$$F_B = \frac{KE_B}{d_B} = \frac{225}{1500} = 0.15 \text{ N}$$
2
$$\frac{F_A}{F_B} = \frac{0.1}{0.15} = \boxed{\frac{2}{3}}$$
$F = KE/d$ | $F_A/F_B = (100/1000)/(225/1500) = 0.1/0.15 = 2/3$
Theory: Laws of Motion
1. Work-Energy Theorem

$W_{net} = \Delta KE = KE_f - KE_i$. Work done by all forces = change in kinetic energy. For braking car (decelerating to stop): $W_{brake} = 0 - \frac{1}{2}mv^2 = -KE_i$. Magnitude of work by brake $= KE_i$. Since $W = F \times d$ (for constant force): $F = KE/d$. Braking force is constant friction force. Shorter stopping distance with same KE → larger braking force. This is the basis of ABS (Anti-lock Braking System): maintains maximum braking force without locking wheels.

2. Kinetic Energy and Stopping Distance

$KE = \frac{1}{2}mv^2$. For uniform deceleration: $v^2 = u^2 - 2as$. At stop: $0 = u^2 - 2as$, so $s = u^2/(2a) = KE/(ma) = KE/F$. Stopping distance $\propto KE/F$. If speed doubles: KE quadruples → stopping distance quadruples (for same force). This is why speed limits matter: 60 km/h vs 120 km/h → stopping distance 4× longer. Highway rule: $d_{stop} \propto v^2$ (not $v$). Reaction distance: $d_{reaction} = v \times t_{reaction}$ (linear in $v$). Total stopping distance = reaction + braking.

3. Power and Efficiency

Power $P = W/t = Fv$. Unit: Watt (W) = J/s. 1 horsepower = 746 W. Efficiency $\eta = $ useful output / total input. Engine power: $P = Fv$ where $F$ = driving force, $v$ = speed. At constant speed: driving force = total resistance (friction + air drag). For car: $P_{min} = F_{resistance} \times v$. Air drag $\propto v^2$: at high speeds, power needed $\propto v^3$. That is why fuel efficiency drops drastically at highway speeds.

4. Conservative and Non-conservative Forces

Conservative: work depends only on initial and final positions, not path. PE can be defined. Examples: gravity, spring force, electrostatic. $W_{conservative} = -(\Delta PE)$. Non-conservative: work depends on path. Cannot define PE. Examples: friction, air drag, normal force on curved path. Energy dissipated as heat. For system with both: $W_{NC} = \Delta KE + \Delta PE = \Delta E_{mechanical}$. Friction (braking force) is non-conservative → converts KE to heat.

5. Elastic and Inelastic Collisions

Elastic: KE conserved. $e = 1$. For equal masses: velocities exchange. Head-on elastic: $v_1^\prime = (m_1-m_2)u_1/(m_1+m_2)$, $v_2^\prime = 2m_1u_1/(m_1+m_2)$. Inelastic: KE not conserved. $e < 1$. Energy lost $= \frac{1}{2}\mu(u_{rel})^2(1-e^2)$ where $\mu = m_1m_2/(m_1+m_2)$. Perfectly inelastic ($e=0$): maximum KE loss. $v_{common} = (m_1u_1+m_2u_2)/(m_1+m_2)$. Momentum always conserved (regardless of collision type).

6. Work Done Against Friction

$W_{friction} = -\mu_k N \times d = -\mu_k mg \cos\theta \times d$. On horizontal: $W_f = -\mu_k mg d$. This work converts to heat. Temperature rise: $\Delta T = W_f/(ms)$ where $s$ = specific heat. Braking: heat goes into brakes and tyres. Race car brakes can reach 900°C! Regenerative braking: converts KE to electrical energy (instead of heat) → stored in battery. Used in electric/hybrid cars: improves efficiency by 20-30%.

7. Variable Force and Work

When force varies: $W = \int F\,dx$ (area under $F$-$x$ graph). Spring: $F = kx$. $W = \frac{1}{2}kx^2$ (stored as elastic PE). For first stretch $x_1$: $W_1 = \frac{1}{2}kx_1^2$. Additional stretch from $x_1$ to $x_2$: $W = \frac{1}{2}k(x_2^2-x_1^2) = \frac{1}{2}k(x_2+x_1)(x_2-x_1) > \frac{1}{2}k(x_2-x_1)^2$. Each additional unit of stretch requires more force and more work (spring gets stiffer effectively). Elastic PE: $U = \frac{1}{2}kx^2$. Maximum at maximum extension.

8. Energy Conservation in Practice

Total mechanical energy $E = KE + PE = $ constant (only for conservative forces). When non-conservative forces act: $E_{final} = E_{initial} - W_{dissipated}$. Roller coaster: $\frac{1}{2}mv^2 + mgh = $ constant (frictionless). Real: speed at bottom slightly less than $\sqrt{2gh}$ (friction losses). Pendulum: $v_{max} = \sqrt{2gL(1-\cos\theta)}$ at bottom. With air resistance: amplitude decreases each swing. Projectile: KE + PE = constant (ignoring air). Maximum height: $KE_0 = mgh_{max}$ for vertical launch.

Frequently Asked Questions
1. Why does F = KE/d come from work-energy theorem? ⌄
Work done by braking force $F$ over distance $d$ = $W = F \times d$ (force and motion in opposite directions, so work is negative). By work-energy theorem: $W = \Delta KE = KE_{final} - KE_{initial} = 0 - KE_i = -KE_i$. Magnitude: $F \times d = KE_i$. Therefore $F = KE_i/d$. This assumes: (1) Constant braking force. (2) Car starts with KE = $KE_i$ and stops ($KE_f = 0$). (3) No other forces doing significant work (horizontal road, ignoring air drag for simplicity).
2. What if both cars were travelling at the same speed? ⌄
If both at speed $v$: $KE_A = \frac{1}{2}m_A v^2 = 100$ J and $KE_B = \frac{1}{2}m_B v^2 = 225$ J. So $m_B/m_A = 225/100 = 2.25$. $F_A = 100/1000 = 0.1$ N, $F_B = 225/1500 = 0.15$ N. Braking deceleration $a = F/m$: $a_A = 0.1/m_A$, $a_B = 0.15/m_B = 0.15/(2.25m_A) = 0.0667/m_A$. Car A decelerates faster ($a_A > a_B$)! Stops faster per unit mass despite smaller force, because it is lighter.
3. What is stopping distance formula in terms of speed? ⌄
For car of mass $m$ moving at speed $v$ with braking force $F$: Deceleration $a = F/m$. Stopping distance $d = v^2/(2a) = mv^2/(2F) = KE/F$. So $d = \frac{1}{2}mv^2/F = KE/F$. This confirms $F = KE/d$. At double speed: $KE$ quadruples, $d$ quadruples (for same $F$). This is the key safety rule: if you double your speed, your braking distance quadruples! At 60 km/h: $d \propto 60^2 = 3600$. At 120 km/h: $d \propto 120^2 = 14400$ (4 times more).
4. How does ABS improve braking? ⌄
ABS (Anti-lock Braking System): prevents wheel lock-up during hard braking. Without ABS: hard braking → wheels lock → slide on road → kinetic friction (lower than static). With ABS: wheel kept just at edge of locking → static friction maintained → maximum braking force. Rolling friction on locked wheel: $f_k = \mu_k mg$. Rolling/static friction: $f_s \leq \mu_s mg$ with $\mu_s > \mu_k$ (typically by 20-30%). ABS also maintains steering control (locked wheels cannot steer). Also reduces stopping distance by 10-20% compared to locked wheels.
5. Can we find individual masses or speeds from this problem? ⌄
No. The problem only gives kinetic energies ($KE = \frac{1}{2}mv^2$) and stopping distances. From $F = KE/d$: $F_A = 0.1$ N, $F_B = 0.15$ N. We cannot determine individual masses or speeds because we only have one equation per car ($\frac{1}{2}mv^2 = $ given) — two unknowns ($m$ and $v$). Additional information would be needed (e.g., deceleration, mass, or speed of one car). The ratio $F_A/F_B$ is determinable because it only requires the given KE and distance values.
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