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).

$N = 98(0.866)$

$\boxed{N = 84.9 \text{ N}}$

(b) Friction force:

$f_k = \mu_k N = (0.25)(84.9)$

$\boxed{f_k = 21.2 \text{ N}}$ (down incline)

(c) Acceleration:

Net force along incline: $F_{net} = F - mg\sin\theta - f_k$

$F_{net} = 80 - (10)(9.8)\sin(30°) - 21.2$

$F_{net} = 80 - 49 - 21.2 = 9.8 \text{ N}$

$a = \frac{F_{net}}{m} = \frac{9.8}{10}$

$\boxed{a = 0.98 \text{ m/s}^2}$ (up incline)