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Physics: Forces and Energy - Contact and Non-contact Forces

Grade 8IB

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

🔑Concepts

A force is a push or pull upon an object resulting from the object's interaction with another object. Forces are vector quantities, meaning they have both magnitude and direction, measured in Newtons (NN).

Contact Forces occur when two objects are physically touching. Examples include Friction (FfF_f), Air Resistance (FdragF_{drag}), Tension (TT), and Normal Reaction Force (FNF_N).

Non-contact Forces act over a distance through a field, without physical contact. Examples include Gravitational Force (WW), Electrostatic Force, and Magnetic Force.

Mass (mm) is a measure of the amount of matter in an object, measured in kilograms (kgkg). It remains constant regardless of location.

Weight (WW) is the force of gravity acting on an object's mass. It changes depending on the gravitational field strength (gg), measured in Newtons (NN).

Normal Reaction Force (FNF_N) is the support force exerted upon an object that is in contact with another stable object, acting perpendicular to the surface.

Resultant (Net) Force (FnetF_{net}) is the single force that represents the combined effect of all forces acting on an object.

📐Formulae

W=m×gW = m \times g

Fnet=m×aF_{net} = m \times a

Fnet=FF_{net} = \sum F

gearth9.8 N/kgg_{earth} \approx 9.8 \text{ N/kg}

💡Examples

Problem 1:

An object has a mass of 25 kg25\ kg. Calculate its weight on Earth where the gravitational field strength gg is 9.8 N/kg9.8\ N/kg.

Solution:

W=25 kg×9.8 N/kg=245 NW = 25\ kg \times 9.8\ N/kg = 245\ N

Explanation:

To find the weight, we multiply the mass of the object by the gravitational acceleration of the planet. Weight is a non-contact force.

Problem 2:

A box is being pushed across a floor with a forward force of 50 N50\ N. The force of friction acting against the motion is 15 N15\ N. Calculate the resultant force.

Solution:

Fnet=50 N15 N=35 N (in the direction of the push)F_{net} = 50\ N - 15\ N = 35\ N \text{ (in the direction of the push)}

Explanation:

Since friction is a contact force acting in the opposite direction to the motion, we subtract it from the applied force to find the net force.

Problem 3:

If a rock weighs 100 N100\ N on Earth (g=9.8 N/kgg = 9.8\ N/kg), what would be its mass on the Moon where g=1.6 N/kgg = 1.6\ N/kg?

Solution:

m=Wg=100 N9.8 N/kg10.2 kgm = \frac{W}{g} = \frac{100\ N}{9.8\ N/kg} \approx 10.2\ kg

Explanation:

Mass is an intrinsic property and does not change based on location. First, calculate the mass on Earth; that mass remains 10.2 kg10.2\ kg on the Moon.

Contact and Non-contact Forces - Revision Notes & Key Formulas | IB Grade 8 Science