PE = mghPE = (3 kg ) x (9.8 m/s/s) x (0.45 m)PE = 13.2 J
Motion and energy
Speed – Distance travelled by any object in unit time is known as speed.Speed – Distance travelled (S.I. unit = m/s) TimeAvg. Speed – Initial speed + Final speed (where acceleration is constant) 2For e.g., -If a man travels 2 km in 20 mins, then he catches a train and coversanother 20 km in 15 mins. Find the avg. speed in its S.I. unit ?Solution –Speed= total distance travelled Time = (2000 + 20000) 1km = 1000m 2100 1min = 60sec = (22000) 2100 = 10.48 m/s
Velocity – It is the distance travelled in unit time in a straight directionor simply speed in a particular direction.Velocity (v) = Distance move in straight direction (SIUnit=m/s) Time TakenDistance travelled on a straight line is known as displacement.Therefore, v = Displacement Time TakenAcceleration – Acceleration is the change in velocity in unit time.Acceleration(a)= Change in velocity (SIUnit=m/s2) Time taken to changeSo, a= v - u (v = final velocity, u = initial velocity, t =
Distance – Time Graph Slope of distance time graph gives speed.If the line is straight the object is travelling with uniform speed or velocity. If the line is parallel to the x-axis the object is at rest.
• Slope of velocity time graph shows acceleration• If line is straight and is making a angle with the x axis there is uniform acceleration• If the line is parallel to the x axis the object is in constant velocity• Area enclosed inside the graph shows the distance
Problems involving uniform acceleration can be solvedquickly using equation of motion.First equation v = final velocity v = u + at u = initial velocity S = distanceSecond equation t = time S = ut + ½ at2 a = accelerationThird equations v2 = u2 + 2aS
Speed 5 2.5 2 4 6 8 10 12 14 Time(m/s)The graph shows a incomplete velocity time graph for a boy running 100 m Find the acceleration during the first 4 seconds a = (v – u) / t = (5 – 0)/4 = 5/4 m/s2 =1.25 m/s2 How far does the boy travel in 4 seconds Distance travelled can be found out by a) equation of acceleration b) by area enclosed in side graph s = [(u + v)/2]t = 2.5 x 4= 10m Other method Area of triangle=1/2x base x height ½ x 4 x 5 = 10m
First LawThe body stays at rest, or if moving it continues to move with uniformvelocity, unless an external force makes it behave differently.For Example – A ball rolling on the ground will never come to a stopif the friction does not act on it in a an opposite direction.Mass And InertiaThe tendency of the object to resist any change in its state of rest ormotion is known as inertia. Its effect is seen when a car stopssuddenly and the occupants lurch forward in an attempt to continuemoving. The larger the mass of the body the larger the inertiatherefore we say mass is the measure of inertia.For example - Its difficult to move Priyansh than to move Gaurav.
Second LawThe rate of change of momentum is directly proportional to the externalforce acting on the body.Which gives the result as:-F=maone newton is defined as the force which gives a mass of 1 kg, an accelerationof 1 m/s2.Weight And GravityThe Weight W of a body is the force of gravity acting on it which gives itacceleration g when it is falling.If a body has mass m then,F = maW = mg ( force = weight, g = acceleration due to gravity)Where g = 9.8 N/kg = 9.8 m/s2.Third LawIf a body A exerts a force on body B, then body B exerts an equal butopposite force on body A. Newton’s 1st law
Kinetic And Potential EnergyThe kinetic energy of an object is the energy which it possesses due to its motion.Ek = ½ x mv2Where Ek = kinetic energy, m = mass, v = velocityThe potential energy of a body is the energy it has because of its position orconfiguration. It is having a very wide meaning as p.e. is of different types for e.g.gravitational p.e., electrostatic p.e., magneto static p.e., chemical p.e. Here whileusing term potential energy we mean by gravitational p.e. only.Ep = mghWhere m = mass, g = 9.8 m/s2, h = given heightA mass m at height h above the ground has p.e. = mgh, when it falls, its velocityincreases and it gains k.e. at the expense of its p.e as followed by law 0fconservation of energy. If it starts from rest and air resistance is negligible its k.e.just before touching the ground equals the p.e. lost by it.Thus,½ x mv2 = mgh
Question: A soccer ball of mass 450 g is travelling at a speed of 20m/s. Howmuch kinetic energy does the soccer ball have?AnswerStep 1 : Identify the quantities involvedThe quantities involved are the mass of the soccer ball (450 g). The speedof the soccer ball (20 m/s) and KE the kinetic energy transferred.Step 2 : Identify the equation neededHere we where asked to find the kinetic energy so we use the equationKE = ½ x mv2. It is important to note that in order to apply this equationall quantities must be in SI units.Step 3 : Convert to SI unitsThe mass of the soccer ball is in g which is not a SI. Converting we have 450 gis 0.45 kg. The speed of the ball is already in SI units.Step 4 : Determine solutionInserting the quantities into the equation we getKE = ½ x mv2= ½ 0.45kg (20m/s)2= 90J