Step | Derivation/Formula | Reasoning |
---|---|---|

1 | Understand action-reaction: \vec{F}_{\text{on ball by pavement}} = – \vec{F}_{\text{on pavement by ball}} | Determine the interaction between the basketball and the pavement using Newton’s third law of motion, which states every action has an equal and opposite reaction. |

2 | Determine force during bounce: \vec{F} = m \vec{a} | Apply Newton’s second law of motion to express the force in terms of the basketball’s mass ( m ) and acceleration ( \vec{a} ). |

3 | Identification of force source | Recognize that the force causing the basketball to bounce up is exerted by the pavement. The pavement pushes against the ball upwards as a reaction to the ball pushing down on it while hitting the pavement. |

4 | Conclusion: Yes, a force is needed. |
The force to make the ball bounce back up is provided by the ground/pavement, according to Newton’s laws of motion, particularly the third law (reaction force). |

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- Statistics

Intermediate

Mathematical

GQ

- Inclines, Linear Forces

Intermediate

Mathematical

GQ

^{2}. Find the friction force impeding its motion. What is the coefficient of kinetic friction?

- Inclines, Linear Forces

Advanced

Mathematical

GQ

A horizontal uniform meter stick of mass 0.2 kg is supported at its midpoint by a pivot point. A mass of 0.1 kg is attached to the left end of the meter stick, and another mass of 0.15 kg is attached to the right end of the meter stick. The meter stick is free to rotate in the horizontal plane around the pivot point. What is the tension in the string supporting the left end of the meter stick?

- Linear Forces, Rotational Motion, Torque

Intermediate

Mathematical

GQ

A 135.0 N force is applied to a 30.0 kg box at 42 degree angle to the horizontal. If the force of friction is 85.0, what is the net force and acceleration? If the object starts from rest, how far has it traveled in 3.3 sec?

- 1D Kinematics, Linear Forces

Intermediate

Mathematical

GQ

A person whose weight is 4.92 × 10^{2} N is being pulled up vertically by a rope from the bottom of a cave that is 35.2 m deep. The maximum tension that the rope can withstand without breaking is 592 N. What is the shortest time, starting from rest, in which the person can be brought out of the cave?

- 1D Kinematics, Linear Forces

Beginner

Mathematical

MCQ

The block is moving horizontally at a constant velocity. There are two applied forces on the object as shown in the image. In which direction is the friction force acting on the object?

- Linear Forces

Beginner

Mathematical

GQ

What is weight of a person who has a mass of 75 kg?

- Linear Forces

Intermediate

Mathematical

GQ

A truck is traveling at 35 m/s when the driver realizes the truck as no breaks. He sees a ramp off the road, inclined at 20°, and decides to go up it to help the truck come to a stop. How far does the truck travel before coming to a stop (assume no friction).

- 1D Kinematics, Linear Forces

Advanced

Mathematical

GQ

A 100 kg person is riding a 10 kg bicycle up a 25° hill. The hill is long and the coefficient of static friction is 0.9. The person rides 10 m up the hill then takes a rest at the top. If she then starts from rest from the top of the hill and rolls down a distance of 7 m before squeezing hard on the brakes locking the wheels. How much work is done by friction to bring the bicycle to a full stop, knowing that the coefficient of kinetic friction is 0.65?

- Energy, Inclines

Advanced

Mathematical

FRQ

A pulley system consists of two blocks of mass 5 kg and 10 kg, connected by a rope of negligible mass that passes over a pulley of radius 0.1 meters and mass 2 kg. The pulley is free to rotate about its axis. The system is released from rest, and the block of mass 10 kg starts to move downwards. Assuming that the coefficient of kinetic friction between the pulley and the rope is 0.2, and neglecting air resistance, determine

- Atwood Machine, Rotational Motion

Yes. It’s the normal force from the ground on the ball.

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Kinematics | Forces |
---|---|

\Delta x = v_i t + \frac{1}{2} at^2 | F = ma |

v = v_i + at | F_g = \frac{G m_1m_2}{r^2} |

a = \frac{\Delta v}{\Delta t} | f = \mu N |

R = \frac{v_i^2 \sin(2\theta)}{g} |

Circular Motion | Energy |
---|---|

F_c = \frac{mv^2}{r} | KE = \frac{1}{2} mv^2 |

a_c = \frac{v^2}{r} | PE = mgh |

KE_i + PE_i = KE_f + PE_f |

Momentum | Torque and Rotations |
---|---|

p = m v | \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 |
---|

F = -k x |

T = 2\pi \sqrt{\frac{l}{g}} |

T = 2\pi \sqrt{\frac{m}{k}} |

Constant | Description |
---|---|

g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |

G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |

\mu_k and \mu_s | Coefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion. |

k | Spring constant, in \text{N/m} |

M_E = 5.972 \times 10^{24} , \text{kg} | Mass of the Earth |

M_M = 7.348 \times 10^{22} , \text{kg} | Mass of the Moon |

M_M = 1.989 \times 10^{30} , \text{kg} | Mass of the Sun |

Variable | SI Unit |
---|---|

s (Displacement) | \text{meters (m)} |

v (Velocity) | \text{meters per second (m/s)} |

a (Acceleration) | \text{meters per second squared (m/s}^2\text{)} |

t (Time) | \text{seconds (s)} |

m (Mass) | \text{kilograms (kg)} |

Variable | Derived SI Unit |
---|---|

F (Force) | \text{newtons (N)} |

E, PE, KE (Energy, Potential Energy, Kinetic Energy) | \text{joules (J)} |

P (Power) | \text{watts (W)} |

p (Momentum) | \text{kilogram meters per second (kgm/s)} |

\omega (Angular Velocity) | \text{radians per second (rad/s)} |

\tau (Torque) | \text{newton meters (Nm)} |

I (Moment of Inertia) | \text{kilogram meter squared (kgm}^2\text{)} |

f (Frequency) | \text{hertz (Hz)} |

General Metric Conversion Chart

Conversion Example

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

Start with the given measurement:

`\text{5 km}`

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

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

Perform the multiplication:

`\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}`

Simplify to get the final answer:

`\boxed{5 \times 10^6 \, \text{mm}}`

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

Pico- | p | 10^{-12} | 0.000000000001 |

Nano- | n | 10^{-9} | 0.000000001 |

Micro- | µ | 10^{-6} | 0.000001 |

Milli- | m | 10^{-3} | 0.001 |

Centi- | c | 10^{-2} | 0.01 |

Deci- | d | 10^{-1} | 0.1 |

(Base unit) | – | 10^{0} | 1 |

Deca- or Deka- | da | 10^{1} | 10 |

Hecto- | h | 10^{2} | 100 |

Kilo- | k | 10^{3} | 1,000 |

Mega- | M | 10^{6} | 1,000,000 |

Giga- | G | 10^{9} | 1,000,000,000 |

Tera- | T | 10^{12} | 1,000,000,000,000 |

- Some answers may be slightly off by 1% depending on rounding, etc.
- Answers will use different values of gravity. Some answers use 9.81 m/s
^{2}, and other 10 m/s^{2 }for calculations. - Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
- Bookmark questions that you can’t solve so you can come back to them later.
- Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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