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Motion is something we observe every day-a car moving on the road, a ball thrown in the air, or a person walking. To understand and describe motion scientifically, we need precise terms and concepts. Displacement is the change in position of an object from its starting point, measured with both magnitude and direction. Velocity is the rate at which an object changes its displacement, which means it also depends on direction. When velocity changes over time, we say the object is accelerating; acceleration measures how fast the velocity changes.
For motion to occur, forces play a vital role. A force is a push or pull on an object that can cause it to start moving, stop, change direction, or change speed. Understanding motion and forces helps us not only in solving physics problems but also in grasping how machines, vehicles, sports, and natural phenomena work.
Let's begin by clarifying important terms:
Types of motion: Uniform motion means the velocity remains constant; the object covers equal distances in equal time intervals. Non-uniform motion means velocity changes-this could be due to speeding up, slowing down, or changing direction.
When acceleration is constant, the motion is said to have uniform acceleration, such as a car steadily speeding up on a straight road.
To analyze such motions mathematically, we use the SUVAT equations, where:
The basic SUVAT equations for uniform acceleration are:
These equations help us find unknown quantities given some known values, such as finding how far a ball travels or how fast it is moving after certain time under acceleration.
Step 1: Identify given values:
Step 2: Use the equation \(v = u + at\).
\[ v = 0 + (2)(5) = 10\, m/s \]Answer: The final velocity of the car after 5 seconds is \(10\, m/s\).
Sir Isaac Newton described how forces cause motion in three fundamental laws that form the foundation of classical mechanics.
An object will stay at rest or continue moving in a straight line with constant speed unless acted upon by a net external force. This means objects resist changes to their motion.
Example: A book resting on a table stays still until you push it. A moving bicycle keeps moving unless brakes or friction slow it down.
Force, mass, and acceleration are related by the formula:
\[ F = ma \]This means the force applied to an object produces acceleration proportional to the force and inversely proportional to the mass.
Example: Pushing a heavier box requires more force to accelerate it than pushing a lighter box.
For every action, there is an equal and opposite reaction. When you push a wall, the wall pushes back with the same force.
In the diagram above, a block experiences several forces:
Friction is a force that opposes motion between two surfaces in contact. It acts parallel to the surface and always opposes the relative movement or attempted movement.
Example: Walking is possible because friction between your shoes and the ground prevents slipping.
Step 1: List given values:
Step 2: Use Newton's second law:
\[ F = ma = 15 \times 4 = 60\, N \]Answer: A force of \(60\, N\) is required.
Work is done when a force causes displacement of an object. The formula for work done is:
\[ W = F \times d \times \cos \theta \]Here, \(F\) is the force magnitude, \(d\) is the displacement, and \(\theta\) is the angle between the force and displacement directions.
Important: Only the component of the force in the direction of motion does work.
Kinetic Energy is energy due to motion, given by:
\[ KE = \frac{1}{2} m v^2 \]Potential Energy is energy due to position, especially height in gravity:
\[ PE = mgh \]where \(g\) is acceleration due to gravity (\(9.8\, m/s^2\)) and \(h\) is height.
The Law of Conservation of Energy tells us that energy cannot be created or destroyed, only transformed from one form to another.
Step 1: Given values:
Step 2: Calculate work done by friction:
\[ W = F \times d \times \cos \theta = 30 \times 10 \times \cos 180^\circ = 30 \times 10 \times (-1) = -300\, J \]Negative work indicates energy is taken away by friction.
Step 3: If the applied force overcomes friction to move the box, work done by applied force to just overcome friction is +300 J.
Answer: Work done against friction is \(300\, J\).
Step 1: Given data:
Step 2: Use equation \( s = ut + \frac{1}{2} a t^2 \):
\[ s = 10 \times 4 + \frac{1}{2} \times 3 \times 4^2 = 40 + \frac{1}{2} \times 3 \times 16 = 40 + 24 = 64\, m \]Answer: The object's displacement after 4 seconds is \(64\, m\).
Step 1: Given values:
Step 2: Use kinetic energy formula:
\[ KE = \frac{1}{2} m v^2 = \frac{1}{2} \times 5 \times 12^2 = 2.5 \times 144 = 360\, J \]Answer: The kinetic energy is \(360\, J\).
Step 1: Given values:
Step 2: Use equation:
\[ v^2 = u^2 + 2as \implies a = \frac{v^2 - u^2}{2s} = \frac{20^2 - 0}{2 \times 50} = \frac{400}{100} = 4\, m/s^2 \]Answer: The acceleration of the body is \(4\, m/s^2\).
Step 1: Given values:
Step 2: Apply \(v = u + at\):
\[ v = 5 + (1.5)(8) = 5 + 12 = 17\, m/s \]Answer: Final velocity is \(17\, m/s\).
Step 1: Given:
Step 2: Use work done formula:
\[ W = F d \cos \theta = 200 \times 3 \times 1 = 600\, J \]Answer: The work done in lifting the box is \(600\, J\).
When to use: When recalling equations quickly during problem solving.
When to use: Avoid calculation mistakes especially under exam time pressure.
When to use: Simplifies calculation for straight-line forces.
When to use: Especially useful in dynamics and friction-related problems.
When to use: Collision or isolated system questions.
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