| Tutorial Topics Mechanics Formula Sheet Motion in one Dimension MechanicsExamples Kinematics equations Motion in Two Dimensions Newton's Laws of Motion Work and Energy Motion in one Dimension We will discuss the motion of objects with constant acceleration along a straight line. Assuming that the object is moving along the x-axis, the displacement is given by
The Average Velocity and the Average Acceleration of the object are given by 
In all the above expressions subscripts f and i stand for final and initial values. Equations of Kinematics (Equations describing the motion with constant acceleration)
When acceleration is constant, velocity verses time graph is a straight line. Average velocity can be written as 
Using equation (3) we can derive the first equation of kinematics.
Equations (5) and (6) can be combined together to obtain the second equation of kinematics 
Combining equations (5) and (7) and eliminating 't' we can obtain 
Find questions and answers in One Dimensional Motion Motion in Two Dimensions Two-dimensional motion can be described as one that takes place in the x-y plane. Imagine a particle moving from point 1 to point 2 in the x-y plane. Let the position vector of these two points be rf and ri. The displacement vector, average velocity and average acceleration are defined as:
Solving Problems in Two-dimension
Kinematics equations can still be used by considering the motion in the x and y directions separately. One example of a two dimensional motion is the projectile motion. Let’s look at an example where an object was thrown at an angle θ with respect to the ground with an initial speed of u. Such an object follows a parabolic path. Lets find the range, time of travel, final speed and the maximum height that the object reach. Projectile motion
1. For simplicity one can assume that the air resistance is zero.
2. Horizontal acceleration is zero (because acceleration of gravity acts vertically downward).
3. Horizontal speed is constant (because the Acceleration = 0)
4. Vertical acceleration is constant (this is equal to the acceleration of gravity, 9.8 m/s2)
Example Consider an object thrown at an angle θ with respect to the ground with an initial speed of u. (a) Find the time of flight (b) Find the range of the projectile (c) Find the maximum height of the that the projectile reached 
Horizontal initial velocity ux = u cos θ
Vertical initial velocity uy = u sin θ
Between point 1 and 3,
Using eq. (7) vertically and horizontally gives the following two expressions:
0 = uy t – ½ g t2 ==> uy = ½ gt ==> t = 2uy/g ==> t = 2 u sin θ / g
s = ux t = u cos θ * 2 u sin θ / g ==> s = (2u2/g) cos θ sin θ
Using eq. (5) horizontally and vertically give the following two expressions:
vx = ux = u cos θ (since a = 0) ==> vx = u cos θ
vy = uy – gt = u sin θ – g(2 u sin θ / g) = - u sin θ ==> vy = - u sin θ
Final velocity v = sqrt (vy2 + vx2)
Between point 1 and 2
Using eq. (7) vertically gives following expression:
0 = uy^2 – 2gHmax ==> Hmax = uy2 /2g = (u2/2g) sin2 θ ==> Hmax = (u2/2g) sin2 θ
Find questions and answers in Two Dimensional Motion
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Newton’s Laws of Motion Newton’s First Law An object at rest continues to be at rest, and an object in motion continues to be in motion with constant velocity as long as the net external force acting on the object is zero. Lets write this statement in mathematical form. Consider an object which is moving with constant velocity. Hence the object's acceleration, 'a' is zero. i.e Fnet = 0 implies that a = 0 |
Newton’s second law
When a net force is acted upon an object, the resulting acceleration is proportional to the net force and inversely proportional to the mass of the object. The acceleration is in the direction of the net force. Mathematically it is written as,
F = m a where F is the net force on the object, m is its mass, and a is its acceleration. F and a are vector quantities.
Force is measured in Newton (N) in SI unit system. |
Newton’s third law
When two objects are in contact, the force exerted by object 1 on object 2 is same in magnitude and opposite in direction to the force exerted by object 2 on object 1.
Newtons’s third law is also phrased as follows:
For every action there is an equal and opposite reaction. 
A note about the diagram: 'mg' is the force of gravity on the object due to the gravitational field of the earth. R is not the reaction force of 'mg'. Rather, R is the reaction force exerted by the surface of contact. |
Example A mass m1 resting on a frictionless table is connected to a cable that passes ver a pulley. The other end of the cable is connected to a mass of m2. Find the acceleration of the masses and the tension in the cable in terms of the given quantities. Assume that there is no friction between the pulley and the cable. Solution 
Forces are as marked in the diagram. Tension on the string is same everywhere since there is no friction on the pulley. Accelerations ‘a’ of the masses are same as well.
Use Newton’s second law to both masses.
For the mass on the table, apply the law horizontally.
T = m1a ------------------------- (1)
For the hanging mass, apply the law vertically.
m2g – T = m2a ------------------------- (2)
(1) + (2) => m2g = (m1 + m2)a
a = m2g / (m1 + m2)
Substituting in (1),
T = m1m2g/(m1 + m2)
Find questions and answers in Newton's Laws of Motion |
Work and Energy Different forms of Energy Energy can appear in different forms. They are mechanical energy, nuclear energy, electromagnetic energy, thermal energy and chemical energy. In this section we will discuss only one type of energy namely mechanical energy. Mechanical energy can present itself as kinetic energy and/or potential energy.
Kinetic Energy
Kinetic energy is related to the motion of an object. An object can undergo different types of motion. They are translational motion, rotational motion or simple harmonic motion (like in a spring mass system). Kinetic energy of a mass undergoing translational motion can be written as,
Kinetic energy = ½ m v2
Potential energy
An example of the potential energy is the gravitational potential energy. An object is said to have a gravitational potential energy if it is in a gravitational field. For example, consider a mass (m) above the surface of the earth. It is in the gravitational field of the earth. For this mass,
Potential energy = mgh
where g is the gravitational acceleration close to the surface of the earth and h is the height of the mass above the earth’s surface. Potential energy is a relative quantity. It is always measured with respect to some reference point in space. Most commonly, the ground is considered the zero potential energy level. Work
Work performed on an object is the product of the magnitude of the displacement and the force on the object along the direction of the displacement. Mathematically, it can be written as,
W = F x r
where F is the force along the direction of the displacement and r is the displacement.
In vector notation, one can write this as,
W = F . r
i.e. the work is the scalar product of two vectors F and r. 
Conservation of Energy
Energy of a particle or a system of particles can change from one form of energy to another. But, in an isolated system, total energy is always conserved. Initial (Kinetic Energy + Potential Energy) = Final (Kinetic Energy + Potential Energy) Consider the following example: Example A block of mass m is released from a top of a frictionless ramp. Find its speed at the bottom of the ramp after the block slides down a vertical distance of ‘h’. Solution 
Its initial speed is zero. Hence its initial kinetic energy is zero. Therefore, it’s initial total energy = the gravitational potential energy = mgh, where h is the height of the release point.
Assume that the speed of the block is v at the bottom of the ramp. At this point, its total energy = kinetic energy = ½ m v2
By conservation of energy,
Total initial energy = Total final energy
m g h = ½ m v2
gh = v2/2
v = sqrt (2gh) |
Force of Friction
Many problems in physics make the assumption of smooth surfaces. However, in real life, surfaces are almost always rough and cannot be considered frictionless. In those situations, one cannot use the conservation of energy concept without considering the energy loss. Two quantities are often used in discussing the force of friction. They are coefficient of static friction, μs and coefficient of kinetic friction, μk.
Static Friction
Static friction exists between two objects that are at rest relative to each other. Consider a block resting on a rough surface. Try applying a very small force. Block may not move. Try to gradually increase the applied force. At one point the block will start to move. When you initially applied a smaller force, static friction was equal to the applied force, hence the block did not move. As you increase the applied force, static friction increases as well reaching a limiting value of F = μsR
Static Frictional Force, F = μsR where R is the normal force.
Kinetic friction
For an object at relative motion with respect to another surface, Frictional Force, F = μkR. Objects moving against frictional force lose some of its energy. Hence mechanical energy is not conserved. In these situations, relationship between initial and final mechanical energies and energy loss due to friction can be given by:
Ei – Ef = Wf = Work done against friction
½ m vi2 + mghi – (½ m vf2 + mghf) = Wf Here i and f denote the initial and final values. Wf is the work done against friction. |
Example A 80 kg baseball player slides into second base when moving at a speed of 5.0 m/s. The coefficient of friction between him and the ground is 0.73. His speed as he reaches the second base is zero. (a) How much mechanical energy is lost due to friction? (b) How far does he slide? 
(a)
Energy lost = Initial kinetic energy - final kinetic energy
= ½ m v^2
= ½ * 80 * 25
= 1000 J (b)
R, in the above diagram is the normal force.
R = mg
Frictional force = F = μR = μ mg
Apply Newton’s second law:
F = m a
μ mg = ma
Acceleration, a = μg
a = 0.73 * 9.8 = 7.15 m/s^2
Here a is the deceleration of the player.
Use v^2 = u^2 + 2a s to find the distance of slide.
0 = 5^2 – 2* 7.154 * s
s = 1.7 m
Distance he slides = 1.7 m
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Example A 15 kg crate is pulled up a rough incline with a force of 150 N applied parallel to the incline, which makes an angle of 30o with the horizontal. Initial speed of the crate is 2.0 m/s. If the coefficient of kinetic friction is 0.47 and the crate is pulled a distance of 6.0 m, (a) how much work is done by the gravitational force? (b) How much work is done by the frictional force (c) what is the work done by the 150 N force and (d) What is the change in kinetic energy of the crate? Find the solution to this problem here (Question #14) |
Consider a ball of radius r and mass M rolling down an incline starting from rest. 
Let Angular acceleration = α Linear acceleration = a Angle of Inclination = θ Height of ball release = h Angular velocity = ω Linear velocity = v Moment of Inertia = I Radius of the ball = r Forces acting on the ball are: Frictional force F Gravitational force mg Reaction force R Energy considerations: 1. Transnational Kinetic Energy = ½ m v2 2. Rotational Kinetic Energy = ½ I ω2 where I is the moment of inertia of the ball with respect to the axis of rotation (diameter of the ball) 3. Potential energy = mgh Using conservation of energy, 0 + 0 + mgh = ½ m v2 + ½ I ω2 mgh = ½ m v2 + ½ I r2 ω2 ------------------ (1) τ = I α = F r ------------------ (2) mg sin θ – F = m a ------------------ (3) v = r ω ------------------ (4) a = α r ------------------ (5) (2) and (3) è mg sin θ – I α/r = m a Using (5), mg sin θ – I a/r2 = m a a = mg sin θ / (m + I/r2) For a sphere, I = 2/5 mr2 a = mg sin θ / (m + 2/5 m) = 5g sin θ / 7 = 5g/7 sin θ a = 5g/7 sin θ ------------------ (6)
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