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Force and Laws of Motion - Forces Acting on a System of Objects

Grade 9CBSE

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

🔑Concepts

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A Force is an external agent capable of changing the state of rest or motion of a particular body. It is a vector quantity with the SI unit Newton (NN).

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Resultant Force (FnetF_{net}): When multiple forces act on a system, the net force is the vector sum of all individual forces. If forces act in the same direction, they are added; if they act in opposite directions, the smaller is subtracted from the larger.

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Balanced Forces: If the resultant of all forces acting on a body is zero (Fnet=0F_{net} = 0), the forces are balanced. Balanced forces do not change the state of rest or motion but can change the shape of an object.

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Unbalanced Forces: If the resultant force is non-zero (Fnet≠0F_{net} \neq 0), the forces are unbalanced. These forces cause a change in the speed or direction of motion, resulting in acceleration.

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Newton's Second Law of Motion: The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force. Mathematically, F=m×aF = m \times a.

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Inertia: The natural tendency of an object to resist a change in its state of motion or rest. The mass (mm) of an object is a measure of its inertia.

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Newton's Third Law: When one object exerts a force on another object, the second object instantaneously exerts a force back on the first. These two forces are equal in magnitude but opposite in direction (FAB=−FBAF_{AB} = -F_{BA}).

📐Formulae

F=m×aF = m \times a

Fnet=F1+F2+F3+...F_{net} = F_1 + F_2 + F_3 + ...

p=m×vp = m \times v

Acceleration (a)=v−ut\text{Acceleration } (a) = \frac{v - u}{t}

1 N=1 kg⋅m/s21 \text{ N} = 1 \text{ kg} \cdot \text{m/s}^2

💡Examples

Problem 1:

A wooden block of mass 5 kg5 \text{ kg} is pulled along a horizontal surface by two forces acting in the same direction: F1=15 NF_1 = 15 \text{ N} and F2=10 NF_2 = 10 \text{ N}. If the friction acting against the motion is 5 N5 \text{ N}, calculate the net force and the acceleration of the block.

Solution:

Total Applied Force: 15 N+10 N=25 N15 \text{ N} + 10 \text{ N} = 25 \text{ N} Frictional Force (opposing motion): 5 N5 \text{ N} Net Force (FnetF_{net}): 25−520\begin{array}{r} 25 \\ - 5 \\ \hline 20 \end{array} So, Fnet=20 NF_{net} = 20 \text{ N}. Using F=maF = ma: a=Fnetm=205=4 m/s2a = \frac{F_{net}}{m} = \frac{20}{5} = 4 \text{ m/s}^2

Explanation:

To find the net force on a system, we add forces acting in the direction of motion and subtract forces acting in the opposite direction (like friction). The resulting force causes the mass to accelerate according to Newton's Second Law.

Problem 2:

Calculate the force required to impart a car of mass 1200 kg1200 \text{ kg} a velocity of 30 m/s30 \text{ m/s} in 10 seconds10 \text{ seconds} starting from rest.

Solution:

Given: Mass (mm) = 1200 kg1200 \text{ kg} Initial velocity (uu) = 0 m/s0 \text{ m/s} Final velocity (vv) = 30 m/s30 \text{ m/s} Time (tt) = 10 s10 \text{ s} Acceleration (aa): a=v−ut=30−010=3 m/s2a = \frac{v - u}{t} = \frac{30 - 0}{10} = 3 \text{ m/s}^2 Force (FF): F=m×a=1200×3=3600 NF = m \times a = 1200 \times 3 = 3600 \text{ N}

Explanation:

First, we determine the acceleration required to reach the target velocity within the given time frame. Then, we apply Newton's Second Law formula (F=maF = ma) to find the total force needed.