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Forces and Motion - Friction and Air Resistance

Grade 9IB

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

🔑Concepts

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Friction is a resistive force that opposes the relative motion or the tendency of motion between two surfaces in contact, acting parallel to the surface interface.

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The magnitude of friction depends on the nature of the surfaces (represented by the coefficient of friction μ\mu) and the normal reaction force RR (or FNF_N).

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Air resistance, or drag, is a type of friction that acts on objects moving through the air. It increases as the object's speed vv and cross-sectional area AA increase.

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

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Static friction fsf_s acts when an object is stationary but an external force is applied. Kinetic (dynamic) friction fkf_k acts when the object is sliding.

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Lubrication and streamlining (aerodynamic shaping) are methods used to reduce friction and air resistance respectively.

📐Formulae

fmax=μRf_{max} = \mu R

Fnet=maF_{net} = ma

W=mgW = mg

Fnet=Fapplied−fF_{net} = F_{applied} - f

R=mgcos⁡(θ) (on an inclined plane)R = mg \cos(\theta) \text{ (on an inclined plane)}

💡Examples

Problem 1:

A wooden crate of mass 20 kg20\text{ kg} is pushed along a horizontal floor with a constant force of 80 N80\text{ N}. If the coefficient of kinetic friction μk\mu_k between the crate and the floor is 0.250.25, calculate the acceleration of the crate. (Assume g=9.8 m/s2g = 9.8\text{ m/s}^2)

Solution:

R=mg=20×9.8=196 NR = mg = 20 \times 9.8 = 196\text{ N} fk=μkR=0.25×196=49 Nf_k = \mu_k R = 0.25 \times 196 = 49\text{ N} Fnet=Fapplied−fk=80−49=31 NF_{net} = F_{applied} - f_k = 80 - 49 = 31\text{ N} a=Fnetm=3120=1.55 m/s2a = \frac{F_{net}}{m} = \frac{31}{20} = 1.55\text{ m/s}^2

Explanation:

First, we find the normal reaction force RR, which equals the weight on a flat surface. Then, we calculate the friction force using f=μRf = \mu R. The net force is the difference between the applied push and friction. Finally, Newton's Second Law F=maF = ma is used to find acceleration.

Problem 2:

A skydiver of mass 70 kg70\text{ kg} is falling through the air. At a specific moment, the air resistance acting on the skydiver is 500 N500\text{ N}. Determine the skydiver's acceleration at this instant.

Solution:

W=mg=70×9.8=686 NW = mg = 70 \times 9.8 = 686\text{ N} Fnet=W−Fair=686−500=186 N (downwards)F_{net} = W - F_{air} = 686 - 500 = 186\text{ N (downwards)} a=Fnetm=18670≈2.66 m/s2a = \frac{F_{net}}{m} = \frac{186}{70} \approx 2.66\text{ m/s}^2

Explanation:

The skydiver is acted upon by weight WW downwards and air resistance FairF_{air} upwards. Since weight is greater than air resistance, the skydiver continues to accelerate downwards, but at a rate less than gg (9.8 m/s29.8\text{ m/s}^2).