A student is watching their hockey puck slide up and down an incline. They give the puck a quick push along a frictionless table, and it slides up a $ 30^\circ $ rough incline ($ \mu_k = 0.4 $) of distance $ d $, with an initial speed of $ 5 $ $ \text{m/s} $, and then it slides back down.(a) Does it take longer to move up the distance $ d $ or back down the distance $ d $? Or does it take the same amount of time? Explain briefly. , (b) Bonus Challenge: Repeat the problem but assume you are not given the initial speed, angle of incline, or $ \mu_k $. :

It takes less time to slide up. This is because there is a greater net force on the way up (gravity and friction work together to slow the block down quicker).

A student is watching their hockey puck slide up and down an incline. They give the puck a quick push along a frictionless table, and it slides up a $ 30^\circ $ rough incline ($ \mu_k = 0.4 $) of distance $ d $, with an initial speed of $ 5 $ $ \text{m/s} $, and then it slides back down. • Nerd Notes

Objective: Determine whether it takes longer for the hockey puck to slide up or down a distance $d$ on an incline with kinetic friction $μ_{k} = 0.4$ , given an initial speed of 5 m/s.

Analysis for Upward Motion:

Step	Formula Derivation	Reasoning
1	[katex]a_{\text{up}} = -g(\sin(\theta) + \mu_k \cos(\theta))[/katex]	Acceleration is due to gravity and kinetic friction opposing the motion.
2	[katex]v^2 = u^2 + 2a_{\text{up}}d[/katex]	Use the kinematic equation for final velocity. v $= 0$ m/s at the top of incline.
3	[katex]t_{\text{up}} = \frac{v – u}{a_{\text{up}}}[/katex]	Time taken $t_{up}$ is found using the kinematic equation for time.
4	Substitute and solve for $t_{up}$ . [katex]u = 5 \text{ m/s}[/katex], [katex]\theta = 30^\circ[/katex], [katex]\mu_k = 0.4[/katex]	Calculate time up.

Analysis for Downward Motion:

Step	Formula Derivation	Reasoning
5	[katex]a_{\text{down}} = g(\sin(\theta) – \mu_k \cos(\theta))[/katex]	Acceleration is due to gravity assisted by friction.
6	[katex]t_{\text{down}} = \sqrt{\frac{2d}{a_{\text{down}}}}[/katex]	Time taken $t_{down}$ using the kinematic equation for constant acceleration.
7	Substitute and solve for $t_{down}$ . [katex]u = 0 \text{ m/s}[/katex] at the peak, [katex]\theta = 30^\circ[/katex], [katex]\mu_k = 0.4[/katex]	Calculate time down.

Calculating the time for both upward and downward motions:

Step	Result
8	[katex] t_{\text{up}} \approx 0.60 \text{ s} [/katex]
9	[katex] t_{\text{down}} \approx 1.15 \text{ s} [/katex]

Step

Result

[katex] t_{\text{up}} \approx 0.60 \text{ s} [/katex]

[katex] t_{\text{down}} \approx 1.15 \text{ s} [/katex]

Kinematics	Forces
\(\Delta x = v_i t + \frac{1}{2} at^2\)	\(F = ma\)
\(v = v_i + at\)	\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)	\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)	\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\)

Circular Motion	Energy
\(F_c = \frac{mv^2}{r}\)	\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)	\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)	\(KE_i + PE_i = KE_f + PE_f\)
	\(W = Fd \cos\theta\)

Momentum	Torque and Rotations
\(p = mv\)	\(\tau = r \cdot F \cdot \sin(\theta)\)
\(J = \Delta p\)	\(I = \sum mr^2\)
\(p_i = p_f\)	\(L = I \cdot \omega\)

Simple Harmonic Motion	Fluids
\(F = -kx\)	\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)	\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)	\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)	\(F_b = \rho V g\)
\(a = -\omega^2 x\)	\(A_1v_1 = A_2v_2\)

Constant	Description
[katex]g[/katex]	Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface
[katex]G[/katex]	Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex]
[katex]\mu_k[/katex] and [katex]\mu_s[/katex]	Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion.
[katex]k[/katex]	Spring constant, in [katex]\text{N/m}[/katex]
[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex]	Mass of the Earth
[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex]	Mass of the Moon
[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex]	Mass of the Sun

Variable	SI Unit
[katex]s[/katex] (Displacement)	[katex]\text{meters (m)}[/katex]
[katex]v[/katex] (Velocity)	[katex]\text{meters per second (m/s)}[/katex]
[katex]a[/katex] (Acceleration)	[katex]\text{meters per second squared (m/s}^2\text{)}[/katex]
[katex]t[/katex] (Time)	[katex]\text{seconds (s)}[/katex]
[katex]m[/katex] (Mass)	[katex]\text{kilograms (kg)}[/katex]

Variable	Derived SI Unit
[katex]F[/katex] (Force)	[katex]\text{newtons (N)}[/katex]
[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy)	[katex]\text{joules (J)}[/katex]
[katex]P[/katex] (Power)	[katex]\text{watts (W)}[/katex]
[katex]p[/katex] (Momentum)	[katex]\text{kilogram meters per second (kgm/s)}[/katex]
[katex]\omega[/katex] (Angular Velocity)	[katex]\text{radians per second (rad/s)}[/katex]
[katex]\tau[/katex] (Torque)	[katex]\text{newton meters (Nm)}[/katex]
[katex]I[/katex] (Moment of Inertia)	[katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex]
[katex]f[/katex] (Frequency)	[katex]\text{hertz (Hz)}[/katex]

Prefix	Symbol	Power of Ten
Pico-	p	[katex]10^{-12}[/katex]
Nano-	n	[katex]10^{-9}[/katex]
Micro-	µ	[katex]10^{-6}[/katex]
Milli-	m	[katex]10^{-3}[/katex]
Centi-	c	[katex]10^{-2}[/katex]
Deci-	d	[katex]10^{-1}[/katex]
(Base unit)	–	[katex]10^{0}[/katex]
Deca- or Deka-	da	[katex]10^{1}[/katex]
Hecto-	h	[katex]10^{2}[/katex]
Kilo-	k	[katex]10^{3}[/katex]
Mega-	M	[katex]10^{6}[/katex]
Giga-	G	[katex]10^{9}[/katex]
Tera-	T	[katex]10^{12}[/katex]

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