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Physics: Motion and Mechanics - Friction, Air Resistance, and Terminal Velocity

Grade 8IB

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

🔑Concepts

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Friction is a contact force that opposes the relative motion or tendency of motion between two surfaces. It acts parallel to the surfaces in contact and in the direction opposite to motion.

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The magnitude of friction depends on the nature of the surfaces (roughness) and the Normal Force (FNF_N), which is the perpendicular force pressing the surfaces together.

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Air Resistance (Drag) is a type of frictional force exerted by air molecules against a moving object. It increases as the object's speed increases and is affected by the object's cross-sectional area and shape (streamlining).

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Terminal Velocity is the constant maximum velocity reached by a falling object when the upward force of air resistance (FairF_{air}) equals the downward force of gravity (Weight, WW).

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At terminal velocity, the net force (FnetF_{net}) is 0 N0 \text{ N}, meaning the object stops accelerating and continues at a constant speed (a=0 m/s2a = 0 \text{ m/s}^2).

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Streamlining involves shaping an object to reduce air resistance by allowing air to flow more smoothly around it, which is critical in the design of cars, planes, and high-speed trains.

📐Formulae

Fnet=m×aF_{net} = m \times a

W=m×gW = m \times g

Fnet=W−FairF_{net} = W - F_{air}

Ffriction=μ×FNF_{friction} = \mu \times F_{N}

At Terminal Velocity: Fair=W\text{At Terminal Velocity: } F_{air} = W

💡Examples

Problem 1:

A skydiver with a mass of 75 kg75 \text{ kg} is falling through the air. At a certain point, the air resistance acting on them is 500 N500 \text{ N}. Calculate the net force acting on the skydiver and their acceleration. (Take g=10 m/s2g = 10 \text{ m/s}^2)

Solution:

  1. Calculate Weight: W=m×g=75 kg×10 m/s2=750 NW = m \times g = 75 \text{ kg} \times 10 \text{ m/s}^2 = 750 \text{ N}.
  2. Calculate Net Force: Fnet=W−Fair=750 N−500 N=250 NF_{net} = W - F_{air} = 750 \text{ N} - 500 \text{ N} = 250 \text{ N} (downwards).
  3. Calculate Acceleration: a=Fnetm=250 N75 kg≈3.33 m/s2a = \frac{F_{net}}{m} = \frac{250 \text{ N}}{75 \text{ kg}} \approx 3.33 \text{ m/s}^2.

Explanation:

The skydiver is still accelerating because the downward force of gravity is greater than the upward force of air resistance.

Problem 2:

A wooden crate is being pushed across a floor with a horizontal force of 120 N120 \text{ N}. If the force of friction between the crate and the floor is 45 N45 \text{ N}, what is the net force acting on the crate?

Solution:

Fnet=Fpush−FfrictionF_{net} = F_{push} - F_{friction} Fnet=120 N−45 N=75 NF_{net} = 120 \text{ N} - 45 \text{ N} = 75 \text{ N}

Explanation:

The net force is the resultant of the applied force and the resistive force of friction acting in the opposite direction.

Problem 3:

A ball reaches terminal velocity while falling. If the ball has a mass of 0.5 kg0.5 \text{ kg}, what is the magnitude of the air resistance acting on it? (Take g=10 m/s2g = 10 \text{ m/s}^2)

Solution:

At terminal velocity, Fnet=0F_{net} = 0, so Fair=WF_{air} = W. W=m×g=0.5 kg×10 m/s2=5 NW = m \times g = 0.5 \text{ kg} \times 10 \text{ m/s}^2 = 5 \text{ N} Therefore, Fair=5 NF_{air} = 5 \text{ N}.

Explanation:

Terminal velocity is defined as the state where the drag force exactly balances the weight of the object, resulting in zero acceleration.