krit.club logo

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

gearth≈9.8 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 N−15 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/kg≈10.2 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.