Title Notes_Work, Power and Energy_Work, Power and Energy.pdf ✅

www.thinkiit.in Page 1 thinkIIT WORK, POWER & ENERGY Work  Defination : In our daily life "work" implies an activity resulting in muscular or mental exertion. However, in physics the term 'work' is used in a specific sense involves the displacement of a particle or body under the action of a force. If under the action of a force F particle or body is displace by S. The scalar product of force with displacement is defined as work. Work is a scalar quantity. W = F .S = FS cos   Another definition of "work is said to be done when the point of application of a force moves. Work done in moving a body is equal to the product of force exerted on the body and the distance moved by the body in the direction of force. Work = Force × Distance moved in the direction of force.  The work done by a force on a body depends on two factors : (i) Magnitude of the force, and (ii) Distance through which the body moves (in the direction of force)  Unit of Work When a force of 1 newton moves a body through a distance of 1 metre in its own direction, then the work done is known as 1 joule. Initial position of the body Final position of the body F A B S Work = Force × Displacement 1 joule = 1 N × 1 m or 1 J= 1 Nm (In SI unit) 1 erg= 1 dyne × 1 cm (In CGS unit) Relationship between joule and erg 1 joule= 1 N × 1 m = 105 dynes × 100 cm ( 1N = 105 dynes, 1 m = 100 cm) 1 joule= 107 dynes × cm = 107 erg 1 joule= 107 erg Ex.1 How much work is done by a force of 10N in moving an object through a distance of 1 m in the direction of the force ? CONTENTS WORK WORK DONE ANALYSIS POWER ENERGY MECHANICAL ENERGY TRANFORMATION OF ENERGY LAW OF CONSERVATION OF ENERGY
www.thinkiit.in Page 2 thinkIIT Sol. The work done is calculated by using the formula : W = F × S Here, Force, F = 10 N And,Distance, S = 1 m So, Work done, W = 10 × 1 J = 10 J Thus, the work done is 10 joules Ex.2 Find the work done by a force of 10 N in moving an object through a distance of 2 m. Sol. Work done= Force × Distance moved Here, Force = 10 N Distance moved = 2 m Work done, W = 10 N × 2 m = 20 Joule = 20 J Work Done Analysis  Work done when force and displacement are along same line.  Work done by a force : Work is said to be done by a force if the direction of displacement is the same as the direction of the applied force.  Work done against the force : Work is said to be done against a force if the direction of the displacement is opposite to that of the force.  Work done against Gravity : To lift an object, an applied force has to be equal and opposite to the force of gravity acting on the object. If 'm' is the mass of the object and 'h' is the height through which it is raised, then the upward force (F) = force of gravity = mg If 'W' stands for work done, then W = F . h = mg . h Thus W = mgh Therefore we can say that, "The amount of work done is equal to the product of weight of the body and the vertical distance through which the body is lifted. Ex.3 Calculate the work done in pushing a cart, through a distance of 100 m against the force of friction equal to 120 N. Sol. Force, F = 120 N; Distance, s = 100 m Using the formula, we have W = Fs = 120 N × 100 m = 12,000 J Ex.4 A body of mass 5 kg is displaced through a distance of 4m under an acceleration of 3 m/s2 . Calculate the work done. Sol. Given :mass, m = 5 kg acceleration, a = 3 m/s2 Force acting on the body is given by F= ma = 5 × 3 = 15 N Now, work done is given by W= Fs = 15 N × 4 m = 60 J
www.thinkiit.in Page 3 thinkIIT Ex.5 Calculate the work done in raising a bucket full of water and weighing 200 kg through a height of 5 m. (Take g = 9.8 ms–2). Sol. Force of gravity mg = 200 × 9.8 = 1960.0 N h = 5 m Work done, W = mgh or W= 1960 × 5 = 9800 J Ex.6 Calculate the work done in lifting 200 kg of water through a vertical height of 6 metres (Assume g = 10 m/s2 ) Sol. In this case work is being done against gravity in lifting water. Now, the formula for calculating the work done against gravity is: W= m × g × h Here, Mass of water, m = 200 kg Acceleration due to gravity, g = 10 m/s2 And, Height, h = 6 m Now, putting these values in the above formula, we get : W = 200 × 10 × 6 W = 12000 J Thus, the work done is 12000 joules.  Work done when force and displacement are inclined (Oblique case) Consider a force 'F' acting at angle to the direction of displacement 's' as shown in fig. F F sin  F cos  A B S  Work done = (component of the force along the direction of motion) × (distance moved) For this we have to first find the component of the applied force along the direction of the displacement.  Resolution of a force into two mutually perpendicular components : Magnitudes of the Components O F A y y ́ B x x ́ F cos F sin  C  Along x-axis :
www.thinkiit.in Page 4 thinkIIT cos = hypotenuse base = OA OB  OB = OA cos = F. cos  Fx = F cos   Along y-axis : sin = hypotenuse opposite = OA AB = OA OC (  OC = AB)  OC = OA . sin = F sin  Fy = F sin   Work done when force is perpendicular to Displacement = 90o W = F.S × cos 90o = F.S × 0 = 0 Thus no work is done when a force acts at right angle to the displacement.  Special Examples :  When a bob attached to a string is whirled along a circular horizontal path, the force acting on the bob acts towards the centre of the circle and is called as the centripetal force. Since the bob is always displaced perpendicular to this force, thus no work is done in this case.  Earth revolves around the sun. A satellite moves around the earth. In all these cases, the direction of displacement is always perpendicular to the direction of force (centripetal force) and hence no work is done.  A person walking on a road with a load on his head actually does no work because the weight of the load (force of gravity) acts vertically downwards, while the motion is horizontal that is perpendicular to the direction of force resulting in no work done. Here, one can ask that if no work is done, then why the person gets tired. It is because the person has to do work in moving his muscles or to work against friction and air resistance.  Important Point : Work can be positive or negative as is acute (< 90o) or obtuse (> 90o). Positive work means that force (or its component) is parallel to displacement while negative works means that force (or its component) is opposite to displacement . Thus  When a person lifts a body from the ground, the work done by the (upward) lifting force is positive, but the work done by the (downward) force of gravity is negative.  Also when a body is pulled on a fixed rough surface, the work done by the pulling force is positive while by frictional force (kinetic) is negative as pulling force is always in the direction while frictional force is opposite to displacement. Ex.7 A boy pulls a toy cart with a force of 100 N by a string which makes an angle of 60o with the horizontal so as to move the toy cart by a distance horizontally. Calculate the work done. Sol. GivenF = 100 N, s = 3 m, = 60o. Work done is given by W = Fs cos = 100 × 2 × cos 60o = 100 × 3 × 2 1 = 150 J ( cos 60o = 2 1 )