0 attempts

0% avg

UBQ Credits

Objective: Determine whether it takes longer for the hockey puck to slide up or down a distance *$d$* on an incline with kinetic friction $μ=0.4$, 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 $tup$ is found using the kinematic equation for time. |

4 | Substitute and solve for $tup $. [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 $tdown $ using the kinematic equation for constant acceleration. |

7 | Substitute and solve for $tdown $. [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] |

The time taken for the hockey puck to move up the distance *$d$* is approximately 0.60 seconds, while the time to slide down the same distance is approximately 1.15 seconds. Therefore, it takes longer for the puck to move down the distance *$d$* than to move up.

This result may seem counterintuitive, but it’s important to note that the initial conditions (starting speed and angle of incline) and the presence of friction significantly influence the motion. The initial upward speed allows the puck to cover the upward distance quickly, while friction continuously slows it down during both upward and downward motion, affecting the total time for each path.

Just ask: "Help me solve this problem."

- Statistics

Intermediate

Conceptual

MCQ

A crate rests on a horizontal surface and a woman pulls on it with a 10-N force. No matter what the orientation of the force, the crate does not move. From least to greatest, rank the normal force on the crate.

- Linear Forces, Tension

Beginner

Mathematical

GQ

A 1100 kg car accelerates from 32 m/s to 8.0 m/s in 4.0 sec. What amount of force was needed to slow it down?

- 1D Kinematics, Linear Forces

Advanced

Mathematical

GQ

A 1.5 kg block is pushed to the right with just enough force to get it to move. The block is pushed for five seconds with this constant force, then the force is released and the block slides to a stop. If the coefficient of kinetic friction is 0.300 and the coefficient of static friction is 0.400. Calculate the amount of time that passes from when the force is applied to when the block stops.

- Linear Forces

Advanced

Mathematical

FRQ

Shown above are three masses of 6 kg, 3 kg, and 1 kg (from left to right). You pull on the first mass with a force of 15 N along a frictionless surface.

- Linear Forces, Multibody Forces

Intermediate

Conceptual

GQ

- Linear Forces

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).

By continuing you (1) agree to our Terms of Sale and Terms of Use and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.

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] |

General Metric Conversion Chart

Conversion Example

Example of using unit analysis: Convert 5 kilometers to millimeters.

Start with the given measurement:

`[katex]\text{5 km}[/katex]`

Use the conversion factors for kilometers to meters and meters to millimeters:

`[katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]`

Perform the multiplication:

`[katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]`

Simplify to get the final answer:

`[katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]`

Prefix | Symbol | Power of Ten | Equivalent |
---|---|---|---|

Pico- | p | [katex]10^{-12}[/katex] | 0.000000000001 |

Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |

Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |

Milli- | m | [katex]10^{-3}[/katex] | 0.001 |

Centi- | c | [katex]10^{-2}[/katex] | 0.01 |

Deci- | d | [katex]10^{-1}[/katex] | 0.1 |

(Base unit) | – | [katex]10^{0}[/katex] | 1 |

Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |

Hecto- | h | [katex]10^{2}[/katex] | 100 |

Kilo- | k | [katex]10^{3}[/katex] | 1,000 |

Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |

Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |

Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |

- 1. Some answers may vary by 1% due to rounding.
- Gravity values may differ: \(9.81 \, \text{m/s}^2\) or \(10 \, \text{m/s}^2\).
- Variables can be written differently. For example, initial velocity (\(v_i\)) may be \(u\), and displacement (\(\Delta x\)) may be \(s\).
- Bookmark questions you can’t solve to revisit them later
- 5. Seek help if you’re stuck. The sooner you understand, the better your chances on tests.

The most advanced version of Phy. 50% off, for early supporters. Prices increase soon.

per month

Billed Monthly. Cancel Anytime.

Trial –> Phy Pro

- Unlimited Messages and Images
- Unlimited UBQ Credits
- 157% Better than GPT
- 30 --> 300 Word Input
- 3 --> 15 MB Image Size Limit
- 1 --> 3 Images per Message
- All Smart Actions
- Mobile Snaps
- Focus Mode
- No Ads

A quick explanation

Credits are used to grade your FRQs and GQs. Pro users get unlimited credits.

Submitting counts as 1 attempt.

Viewing answers or explanations count as a failed attempts.

Phy gives partial credit if needed

MCQs and GQs are are 1 point each. FRQs will state points for each part.

Phy customizes problem explanations based on what you struggle with. Just hit the explanation button to see.

Understand you mistakes quicker.

Phy automatically provides feedback so you can improve your responses.

10 Free Credits To Get You Started