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Forces and Motion - Friction and Air Resistance

Grade 7IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Friction is a contact force that acts between two surfaces sliding, or trying to slide, across each other. It always acts in the direction opposite to the direction of motion.

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The magnitude of friction depends on two main factors: the nature of the surfaces in contact (roughness) and the normal force (FNF_N) pressing the surfaces together.

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Air resistance, also known as drag, is a type of friction that occurs when an object moves through the air. It is caused by the collision of the object's surface with air molecules.

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The size of air resistance depends on the object's speed, its cross-sectional area, and its shape. Streamlining is the process of shaping an object to reduce drag.

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Terminal velocity is reached when the downward force of gravity (WW) is exactly balanced by the upward force of air resistance (FdragF_{drag}), resulting in a net force of 0 N0\text{ N} and zero acceleration (a=0 m/s2a = 0\text{ m/s}^2).

📐Formulae

Fnet=Fapplied−FfrictionF_{net} = F_{applied} - F_{friction}

W=m⋅gW = m \cdot g

a=Fnetma = \frac{F_{net}}{m}

At terminal velocity: Fdrag=W\text{At terminal velocity: } F_{drag} = W

💡Examples

Problem 1:

A wooden crate with a mass of 20 kg20\text{ kg} is pushed across a floor with an applied force of 100 N100\text{ N}. If the force of friction between the crate and the floor is 30 N30\text{ N}, calculate the net force (FnetF_{net}) and the resulting acceleration (aa). Assume g=9.8 m/s2g = 9.8\text{ m/s}^2.

Solution:

Fnet=100 N−30 N=70 NF_{net} = 100\text{ N} - 30\text{ N} = 70\text{ N}

a=70 N20 kg=3.5 m/s2a = \frac{70\text{ N}}{20\text{ kg}} = 3.5\text{ m/s}^2

Explanation:

The net force is the difference between the applied force and the friction opposing it. Using Newton's Second Law (F=maF = ma), we find the acceleration by dividing the net force by the mass.

Problem 2:

A skydiver of mass 70 kg70\text{ kg} is falling through the air. At a certain point, the air resistance acting on the skydiver is 700 N700\text{ N}. If g=10 m/s2g = 10\text{ m/s}^2, what is the skydiver's acceleration?

Solution:

First, calculate the weight (WW): W=70 kg×10 m/s2=700 NW = 70\text{ kg} \times 10\text{ m/s}^2 = 700\text{ N}

Calculate Net Force: Fnet=W−Fdrag=700 N−700 N=0 NF_{net} = W - F_{drag} = 700\text{ N} - 700\text{ N} = 0\text{ N}

Therefore: a=0 m/s2a = 0\text{ m/s}^2

Explanation:

Since the weight (downward force) and air resistance (upward force) are equal and opposite, the net force is zero. The skydiver has reached terminal velocity and is no longer accelerating.