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Forces and Energy - Forces of Motion: Friction, Gravity, and Push/Pull

Grade 4IB

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

🔑Concepts

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A force is a push or a pull acting upon an object as a result of its interaction with another object. Forces are measured in Newtons (NN).

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Friction is a force that resists the relative motion of solid surfaces, fluid layers, and material elements sliding against each other. It acts in the opposite direction of the motion.

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Gravity is a non-contact force that pulls objects toward the center of the Earth. The weight of an object is the gravitational force acting on its mass (mm).

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Forces can change the state of motion of an object. This includes starting motion, stopping motion, or changing the direction of motion.

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Balanced forces occur when the net force (FnetF_{net}) is 0 N0\,N, meaning there is no change in the object's motion. Unbalanced forces cause an object to accelerate or decelerate.

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The mass of an object is the amount of matter in it, while weight is the force of gravity (gg) acting on that mass.

📐Formulae

F=m×aF = m \times a

W=m×gW = m \times g

Fnet=F1+F2+...+FnF_{net} = F_1 + F_2 + ... + F_n

Ffriction=μ×FnormalF_{friction} = \mu \times F_{normal}

💡Examples

Problem 1:

If a student pushes a heavy book across a desk with a force of 15 N15\,N and the friction between the book and the desk is 5 N5\,N, what is the net force acting on the book?

Solution:

10 N10\,N

Explanation:

To find the net force (FnetF_{net}), we subtract the force of friction from the push force because they act in opposite directions: 15 N−5 N=10 N15\,N - 5\,N = 10\,N.

Problem 2:

Calculate the weight (WW) of a toy car on Earth if its mass is 0.5 kg0.5\,kg. (Assume gravity g≈9.8 m/s2g \approx 9.8\,m/s^2)

Solution:

4.9 N4.9\,N

Explanation:

Using the formula W=m×gW = m \times g, we multiply the mass of the car by the acceleration due to gravity: 0.5 kg×9.8 m/s2=4.9 N0.5\,kg \times 9.8\,m/s^2 = 4.9\,N.

Problem 3:

A group of students is playing tug-of-war. Team A pulls with a force of 100 N100\,N to the left, and Team B pulls with a force of 100 N100\,N to the right. What is the state of motion?

Solution:

Fnet=0 NF_{net} = 0\,N

Explanation:

Because the forces are equal in magnitude but opposite in direction (100 N−100 N=0 N100\,N - 100\,N = 0\,N), the forces are balanced and the rope does not move.