Kinetic friction: $f_k = \mu_k N$

where $N$ is the normal force and typically $\mu_s > \mu_k$.

Motion with Friction

For an object sliding with kinetic friction:

$ma = F_{applied} - f_k = F_{applied} - \mu_k mg$

$\frac{dv}{dt} = \frac{F_{applied}}{m} - \mu_k g$

Stopping Distance

Object sliding on horizontal surface with initial velocity $v_0$:

$a = -\mu_k g$

$v\frac{dv}{dx} = -\mu_k g$

$\int_{v_0}^0 v \, dv = -\mu_k g \int_0^d dx$

$-\frac{1}{2}v_0^2 = -\mu_k gd$

$d = \frac{v_0^2}{2\mu_k g}$

Inclined Planes

For angle $\theta$ from horizontal:

Coordinate system: x-axis along incline (positive down), y-axis perpendicular

Weight components:

• Parallel to incline: $F_{\parallel} = mg\sin\theta$
• Perpendicular: $F_{\perp} = mg\cos\theta$

Normal force: $N = mg\cos\theta$ (when no vertical acceleration)

Sliding Down Incline

With kinetic friction:

$ma = mg\sin\theta - \mu_k mg\cos\theta$

$a = g(\sin\theta - \mu_k\cos\theta)$

Condition for sliding: $\sin\theta > \mu_k\cos\theta$, or $\tan\theta > \mu_k$

Velocity after distance $d$: $v^2 = v_0^2 + 2ad = v_0^2 + 2gd(\sin\theta - \mu_k\cos\theta)$

Critical Angle for Static Friction

Object on the verge of sliding:

$f_s = \mu_s N$

$mg\sin\theta_c = \mu_s mg\cos\theta_c$

$\tan\theta_c = \mu_s$

Motion with Time-Dependent Force

Force $F(t)$ applied up an incline:

$m\frac{dv}{dt} = F(t) - mg\sin\theta - \mu_k mg\cos\theta$

$v(t) = v_0 + \int_0^t \frac{F(t')}{m} \, dt' - gt(\sin\theta + \mu_k\cos\theta)$

Example: Exponential Force

If $F(t) = F_0e^{-t/\tau}$:

$v(t) = v_0 + \frac{F_0\tau}{m}(1 - e^{-t/\tau}) - gt(\sin\theta + \mu_k\cos\theta)$

Blocks Connected on Incline

Two blocks (masses $m_1$, $m_2$) connected by rope over pulley:

Block 1 on incline at angle $\theta$, block 2 hanging vertically.

Constraint: $a_1 = a_2 = a$ (magnitude)

Block 1 (up incline positive): $m_1a = T - m_1g\sin\theta - f_k$

Block 2 (down positive): $m_2a = m_2g - T$

Adding equations (T cancels): $a = \frac{m_2g - m_1g\sin\theta - \mu_k m_1g\cos\theta}{m_1 + m_2}$

Energy Considerations with Friction

Work done by friction: $W_f = -f_k \cdot d = -\mu_k mgd\cos\theta$

(negative because friction opposes motion)

This work is converted to thermal energy (non-conservative force).

$a = g(\sin\theta - \mu_k\cos\theta)$

$a = 9.8(\sin 20° - 0.15\cos 20°)$

$a = 9.8(0.342 - 0.15 \times 0.940)$

$a = 9.8(0.342 - 0.141) = 9.8(0.201)$

$\boxed{a = 1.97 \text{ m/s}^2}$

(b) Speed at bottom:

Using $v^2 = v_0^2 + 2aL$:

$v^2 = 0 + 2(1.97)(10) = 39.4$

$\boxed{v = 6.28 \text{ m/s}}$

(c) Energy dissipated:

Friction force: $f_k = \mu_k mg\cos\theta = (0.15)(5.0)(9.8)(0.940) = 6.91 \text{ N}$

Energy dissipated: $E_{friction} = f_k \cdot L = (6.91)(10)$

$\boxed{E_{friction} = 69.1 \text{ J}}$

Check with energy: Height: $h = L\sin\theta = 10(0.342) = 3.42$ m

$\Delta PE = mgh = (5.0)(9.8)(3.42) = 168 \text{ J}$

$KE = \frac{1}{2}mv^2 = \frac{1}{2}(5.0)(6.28)^2 = 98.6 \text{ J}$

$E_{friction} = \Delta PE - KE = 168 - 98.6 = 69.4 \text{ J}$ ✓

where $a_{1x} = a_2 - a_1\cos\theta$ (acceleration of block in lab frame)

For block, vertical components: $0 = N\cos\theta - m_1 g$

From second equation: $N = \frac{m_1 g}{\cos\theta}$

For wedge: $m_2 a_2 = -N\sin\theta$

Combining: $m_1(a_2 - a_1\cos\theta) = \frac{m_1 g\sin\theta}{\cos\theta}$

$m_2 a_2 = -\frac{m_1 g\sin\theta}{\cos\theta}$

From wedge equation: $a_2 = -\frac{m_1 g\tan\theta}{m_2} = -\frac{(2.0)(9.8)\tan 30°}{3.0}$

$a_2 = -\frac{19.6(0.577)}{3.0} = -3.77 \text{ m/s}^2$

$\boxed{a_2 = 3.77 \text{ m/s}^2 \text{ (to left)}}$

(b) Acceleration relative to wedge:

From block equation: $a_2 - a_1\cos\theta = g\tan\theta$

$-3.77 - a_1(0.866) = (9.8)(0.577) = 5.65$

$a_1 = \frac{-3.77 - 5.65}{0.866} = -10.9 \text{ m/s}^2$

$\boxed{a_1 = 10.9 \text{ m/s}^2 \text{ (down incline)}}$

(c) Normal force:

$N = \frac{m_1 g}{\cos\theta} = \frac{(2.0)(9.8)}{\cos 30°}$

$N = \frac{19.6}{0.866}$

$\boxed{N = 22.6 \text{ N}}$