Physics Notes - Relative Motion & Projectile Motion

**DESCRIBING MOTION USING THE CONCEPT OF RELATIVE VELOCITIES IN 2-D** **Relative Motion Velocity** Refers to an object which is relative to some other object that might be stationary, moving with the same velocity, or moving slowly, moving with higher velocity or moving in the opposite direction....

**DESCRIBING MOTION USING THE CONCEPT OF RELATIVE VELOCITIES IN 2-D** **Relative Motion Velocity** Refers to an object which is relative to some other object that might be stationary, moving with the same velocity, or moving slowly, moving with higher velocity or moving in the opposite direction. Basically, this is a matter of perspectives---of how one motion of an object is seen and analyzed in a different angle. **Projectile Motion** - An object that is thrown in a **curve path** with the effect of gravitational field involves **[motion in 2 dimensions]**. This motion is called **projectile** **motion.** - It is motion in **two directions [(]***horizontal and vertica**l*****)** that are **independent** of each otherand with the vertical motion --gravity -- the only force that is acting on. **PROJECTILE --** the object was thrown **TRAJECTORY --** projectile 's path With the effect of gravity, the projectile will accelerate downward and it is equal to the acceleration due to gravity which is 9.8 m/s2. Hence, the acceleration in the **y-direction** is equal to **-9.8 m/s²**. In the **x-direction**, since the force of gravity does not act along this axis, it is equal to **zero (0**). **[INITIAL CONDITIONS]** [The initial conditions are the initial velocity,] 8[V˳ and the angle, Ɵ.] [For example:] [V˳ = 40.0 m/s] [Ɵ = 35°]  **[Components of initial velocity ]** 1. X-component of the initial velocity: **Vox = V˳cosƟ** 2. Y-component of the initial velocity: **Voy = V˳sinƟ** A. What is the time at the maximum height of the trajectory? B. What is the maximum height reached? C. What is the total time of flight? D. What is the range of the projectile? **A.**What is the **Time at the Maximum Height** of the Trajectory? At the maximum height, the velocity along the **y-component is zero** while its velocity is going toward the horizontal direction. At the maximum height, the time is called **the half time of flight.** **B.** What is the **Maximum Height?**. To determine the **maximum height,** the displacement formula is used. **C.** What is the **Total Time of Flight?** From the half time flight (2.33 s), it can be doubled to find the value of the total time of flight. T = 2(t) = 2(2.33 s) T = 4.66 s **T = 4.7 s, the total time of flight** **D.** What is the **Range** of the Projectile? The distance from the initial point on the ground to the final point it reaches is called the **range of the trajectory.** With the effect of gravity, the projectile will accelerate downward and it is equal to the acceleration due to gravity which is 9.8 m/s2. Hence, the acceleration in the **y-direction** is equal to **-9.8 m/s²**. In the **x-direction,** since the force of gravity does not act along this axis, it is equal to **zero (0)**. **NOTE:** Since the **[velocity in the x-direction]** is constant, it maintains its initial velocity throughout the entire flight. **Newton's First Law** **(** Law of Inertia**)** \"An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.\" **Sample Problem** Suppose you are helping a friend transfer from one residence to another and your friend asks you to stand in the rear of a truck to hold a piano from an inclined plane. Explain why you should decline the request of your friend. The answer should indicate your understanding of Galileo and the First Law of Newton. **Answer:** Since a piano has a large mass, it also has a large inertia. If the truck suddenly moves, the piano that is in a state of rest could slide down. Even if the truck stops all of a sudden, the piano will keep moving. Hence, anyone standing in the rear of the piano and the truck could be injured. **Free-Body Diagram** When analyzing physics problems, it is helpful to draw a diagram with all the acting on an object. Such diagram is called a **free-body diagram**. Sample Problem From the following illustrations, verify the magnitude and direction of the net force and recognize which will be at rest.  **Newton's Second Law** **(**Law of Acceleration **)** Once a force that acts on an object is unbalanced, it will accelerate and can be expressed as: a∝ (is proportional to) F, where mass, m, is a constant value When the force remains constant, the mass increases and the acceleration decreases. Mathematically, the relationship can be expressed as: **a ∝ 1/m , where F is constant** The following ideas can be summarized to create the second law of Newton: - Once a force that acts on an object is unbalanced, it will accelerate in the same direction as the force. - The acceleration of the object varies with its unbalanced force. - The acceleration of the object varies inversely with its mass. If the two proportionality statements are combined to form the third proportionality, the result is: **a = F/m** Recall that Newton (symbol N) is the unit of force. 1 N is one kilogram. Meter per second squared. **1 N=1 kg-m/s²** Mathematically, the second law of Newton is: **F=ma** **Sample Problems** 1. If a 70 kg skater acted upon by an unbalanced force of 161 N \[West\], what is its acceleration? 2. A force of 360 N \[East\] was applied by a student on a box with a mass of 50 kg. What is the acceleration of the box if a frictional force of 340 N acts in the opposite direction? Make a free-body diagram to answer the question. Sketch a rectangle to denote the box being analyzed. The 360 N applied force is indicated as a vector acting on the box in the East direction while the frictional force is acting in the opposite direction. 3. A 10 kg box is being pushed by an applied force and accelerates at 2.5 m/s². What is the applied force if a frictional force of 50 N is acting in the opposite direction? M = 10 kg a = 2.5 m/s² F=ma =(10 kg)(2.5 m/s²) =25 N **Using the Free-Body Diagram**  where F, is applied force F, is the force of friction **Newton's Third Law** When an empty balloon is blown with air and released, the air blasts out from the open end of the balloon initiating a movement in the opposite direction. The motion of the balloon is an example of the third law of Motion of For every force of action, there is a force of reaction equal in magnitude but opposite in direction. In the previous example, it can be expressed in an equation as: The force of reaction is when the air is pushing the balloon forward while the escaping air is moving in the opposite direction. **Other Examples of Newton's Third Law** - **Walking** - **Driving** - **Swimming** **Work** is a scalar quantity and is described only by its magnitude. It is simply defined as the dot product of the force and the displacement. 1. There is force applied on an object. 2. The object moves to a distance d as the force is applied. 3. The force applied has a parallel component with the object's motion. 1. What is the work done in pulling a crate 20m horizontally when a Force of 60N is applied on a rope which makes an angle 30" with the ground?  Solution: W= Fdcosθ(60N) (20m) (cos30°) = 1039.2/ 2. How much work is done when a 2.5-kg package is pulled to a distance of 2m along a level floor? (The coefficient friction is 0.2). To calculate the work done when pulling a package, we need to use the formula for work **W=F d cos (θ)** Where : - W is the work done. - F is the force applied, - d is the displacement, - θ is the angle between the force and the direction of displacement. Since the package is being pulled along a level floor, and assuming the force applied is horizontal Ge, 801, the cosine term becomes 1. Thus, the formula simplifies to: **W = F d** **Step 1: Determine the force of friction** The work done in pulling the package is primarily to overcome the force of friction. The force of friction Ff is given by the formula: Ff = μ. Ν Where: μ = 0.2 is the coefficient of friction, N is the normal force, which for a horizontal surface is equal to the weight of the package. The weight of the package is given by: N = mg Where: m = 2.5 kg is the mass of the package, g = 9.8 m/s² is the acceleration due to gravity. Thus, the normal force N is: = 2.5kg x 9.8 m/s² = 24.5 N Now, the force of friction is: Ff = 0.2 x 24.5 N = 4.9 N **Step 2: Calculate the work done** The work done to overcome the friction is: W = Ff d = 4.9 N x 2m = 9.8J Final Answer: The work done to pull the 2.5-kg package a distance of 2 meters along a level floor is 9.8 joules. 3. A factory worker pushes horizontally a 30-kg crate to a distance of 4.5 m along a level floor at a constant velocity. The coefficient of kinetic friction between the crate and the floor is 0.25. A. What magnitude of force (Fw) must the worker apply? B. How much work is done on the crate by this force? C. How much work is done on the crate by the friction force? D. D. How much work is done by the normal force? By the gravity? E. E. What is the net work done on the crate? Given: - Mass of the crate: m 30 kg - Distance moved: d = 4.5 m - Coefficient of kinetic friction: μk = 0.25 - The crate is moving at constant velocity, so the net force is zero. - Gravitational acceleration: g = 9.8 m/s² Step 1: **Determine the force required to move the crate (part a)** Since the crate is moving at constant velocity, the applied force (Fw) must exactly counterbalance the frictional force (Ff) that opposes the motion. The frictional force is given by: F = μk Ν Where: - N is the normal force, and for a horizontal surface, N= mg (since the crate is not accelerating vertically). Now, calculate the normal force: Then, the frictional force is: Ff = μk Ν = 0.25 x 294 N 73.5 N Since the crate is moving at constant velocity, the worker must apply a horizontal force equal in magnitude but opposite in direction to the frictional force: Fw = Ff =73.5 N Step 2: **Work done by the worker's force (part b)** The work done by a force is given by: W = F d cos(θ) In this case, the applied force and the displacement are in the same direction (horizontal), so θ = 0 and cos(0°) = 1. Therefore: Ww = Fw d = 73.5 N x 4.5 m = 330.75 J Answer (b): The work done on the crate by the worker's force is 330.75 J. Step 3: **Work done by the friction force (part c)** The work done by friction is also calculated using the work formula: Wf = Ff d cos(θ) Here, the friction force is acting in the opposite direction to the displacement, so θ= 180°, and cos(180°) = 1. Therefore: Wf = Ff d (-1) = 73.5Nx4.5m = -330.75 J Answer (c) : The work done on the crate by the friction force is **-330.75 J** (negative because friction opposes the motion) Step 4: **Work done by the normal force and gravity (part d)** Both the normal force and gravity act vertically, while the displacement is horizontal. Since the angle between the direction of the force and the direction of displacement is 90°, we have: W = F d cos(90°)=0 Therefore, no work is done by the normal force or gravity. Answer (d): The work done by the normal force is **0 J**. Step 5: **Net work done on the crate(parte)** The net work done on the crate is the sum of all the individual works. We already calculated: - Work done by the worker's force: Ww = 330.75 J - Work done by the friction force: Wf = 330.75 J - Work done by the normal force: Wnormal = OJ - Work done by gravity: Wgravity = 0 J Thus, the total or net work is: Wnet = Ww + Wf + Wnormal + Wgravity = 330. 75 J + (- 330.75) + 0 + 0 = 0 J Answer (e) : The net work done on the crate is 0 J. This is expected because the crate is moving at constant velocity, meaning there is no change in kinetic energy, and thus the net work must be zero (from the work-energy theorem), **ENERGY:** **Potential and Kinetic** **Energy** - The capacity to do work **Potential Energy** - The **stored energy** of position possessed by an object.  **Gravitational Potential Energy** - Is the **energy stored** in an object due to its **position above the Earth's surface.**  **Ep = mgh** m = mass (kg) g = gravitational field strength (N/kg) h = height (m) Ep = gravitational potential energy (J) **Sample Problem** 1. How much gravitational potential energy a 4.0 kg block has if it is lifted 25 m? **ΔEg = mgΔh** =(4.0 kg)(9.80 N/kg)(25 m) = 9.8 x 102 J **Elastic Potential Energy** - Is the energy stored as a result of **deformation of an elastic object**, for example the stretching of a spring.  **Kinetic Energy** Formula**: Ek = ½ mv²** - Ek = Kinetic energy - m = mass - v = velocity  What Is Kinetic Energy? Kinetic Energy Examples  **Summary**