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Forces and Motion - Measuring Force (Newtons)

Grade 7IB

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

πŸ”‘Concepts

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A force is defined as a push or a pull acting upon an object resulting from the object's interaction with another object. Force is measured in Newtons (NN).

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Force is a vector quantity, meaning it has both a magnitude (size) and a specific direction. It can change an object's speed, direction, or shape.

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The instrument used to measure force is called a Newton meter or a spring balance. It works based on the principle that the stretch of the internal spring is proportional to the force applied.

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Mass (mm) is the amount of matter in an object and is measured in kilograms (kgkg), whereas Weight (WW) is the force of gravity acting on that mass and is measured in Newtons (NN).

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Gravitational field strength (gg) represents the force of gravity per unit mass. On Earth, gβ‰ˆ9.8 N/kgg \approx 9.8\,N/kg, though it is often rounded to 10 N/kg10\,N/kg for introductory calculations.

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Balanced forces occur when the resultant force is 0 N0\,N, meaning the object remains at rest or continues at a constant velocity. Unbalanced forces cause acceleration.

πŸ“Formulae

W=mΓ—gW = m \times g

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

F=kΓ—Ξ”xF = k \times \Delta x

πŸ’‘Examples

Problem 1:

Calculate the weight of a science textbook with a mass of 0.5 kg0.5\,kg on Earth, where the gravitational field strength is g=9.8 N/kgg = 9.8\,N/kg.

Solution:

W=0.5 kgΓ—9.8 N/kg=4.9 NW = 0.5\,kg \times 9.8\,N/kg = 4.9\,N

Explanation:

To find the weight, we multiply the mass of the object by the acceleration due to gravity (gg). The resulting unit is Newtons (NN).

Problem 2:

An astronaut weighs 120 N120\,N on the Moon, where the gravity is 1.6 N/kg1.6\,N/kg. What is the astronaut's mass?

Solution:

m=Wg=120 N1.6 N/kg=75 kgm = \frac{W}{g} = \frac{120\,N}{1.6\,N/kg} = 75\,kg

Explanation:

By rearranging the formula W=mΓ—gW = m \times g to solve for mass, we divide the weight by the local gravitational field strength. Mass remains constant regardless of location.

Problem 3:

A spring in a Newton meter has a spring constant of k=50 N/mk = 50\,N/m. If the spring extends by Ξ”x=0.1 m\Delta x = 0.1\,m when an object is hung from it, what is the force being measured?

Solution:

F=50 N/mΓ—0.1 m=5 NF = 50\,N/m \times 0.1\,m = 5\,N

Explanation:

Using Hooke's Law, the force exerted is the product of the spring constant and the extension of the spring.

Measuring Force (Newtons) - Revision Notes & Key Formulas | IB Grade 7 Science