WORK, ENERGY & POWER Work scalar W = F S Cosθ Unit of work in SI system Work done by a constant force

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1 WORK, ENERGY & POWER Work Let a force be applied on a body so that the body gets displaced. Then work is said to be done. So work is said to be done if the point of application of force gets displaced. work is also defined as the product of displacement and the force in the direction of displacement. Work is a scalar quantity. Consider a boy pulling a toy car and walking. The direction of force is along the string. The car moves along the horizontal surface. Let θ be the angle between the direction of force and the horizontal surface. The displacement is caused by the horizontal component of force F (i.e. forces in the direction of displacement) and not by the entire force F. The horizontal component of force F is Fcos θ. Work is defined as the product of displacement and the force in the direction of displacement. If S is the displacement, work done by the force, W = S x F Cosθ. W = F S Cosθ. [Where θ is the angle between the direction of force and direction of displacement]. i.e, work done, W =F. S ie., work done is psitive. (i).if θ =0, W = F S, the maximum work done by the force. (ii).if θ = 90 0, W = F S cos 90 = 0. No work is done. If the force and (iii).when θ=180, i.e. force and displacement are in opposite direction, work done, W = F S cos 180 = - F S. ie work done is negative. Eg: A stone is moving vertically upwards. The gravitational force acts vertically downwards. Here the work done by the gravitational force is negative. Unit of work in SI system Work done, W = F S cosθ. When F = 1 N, S = 1 m and θ= 0, then W = 1 N m. This is called 1 joule. The work done is said to be one joule if a force of one newton can displace a body through one metre in the direction of force. The dimensional formula is ML 2 T -2 Work done by a constant force If the force F is constant, the force-displacement graph is a straight line parallel to the displacement axis(x-axis). Work done = F x s = OA X OC = Area of rectangle.

2 Work done by a variable force Consider a continuously varying force F(x) acting on a body and produces a very small displacement x each time. Since displacement is small, the force F(x) can be considered as constant. Then work done, W = F(x) x. Now a graph drawn connecting F(x) and displacement of body is as shown. Now area of shaded portion, A = F(x) x = Work done W So the total work done in displacing the body from x i to x f, xf xf W= xi F(x). x = W = F(x). x = the total area under the graph. xi Work done in lifting a body:- Let a body of mass m be lifted vertically through a small height h from the surface of the earth then, Force applied to lift the body F = mg Work done in lifting the body = W = F S = mg h. Conservative Force A force is said to be conservative if the work done by the force is independent of the path but depends only on the initial and final positions. *The work done by the conservative force in a closed path (AOBC) is zero. e.g., Gravitational force and elastic spring force are conservative forces. Non-Conservative Force:- If the work done by a force on a particle moving between two points depends on the path taken, the force is called non conservative force. *The total work done by a non-conservative force along a closed path is not zero. e.g., frictional force. (the longer the path greater is the work done). Power Power is defined as work done per unit time. If an agency performs a work W in a time t, then P = W t, Unit of power is joules/sec or Watt. [ 1kW = 1000W, 1 Horse power (H.P) =746W] *Also we have work = FS, then P = W t = FS t = F S t = F x v. (Where v- velocity)

3 Energy :- Energy is the capacity to do work. Energy is also a scalar quantity. Unit is joule (J). There are two types of energy - Kinetic Energy and Potential Energy. 1.Kinetic energy(k.e) The energy possessed by a body by virtue of its motion is called kinetic energy. If a body of mass m moves with a velocity v, then kinetic energy, K= ½ mv 2 *K.E is a scalar quantity and is never negative. Expression for kinetic energy:- Consider a body of mass m moving in a straight line with a velocity v. The K.E of the body must be equal to the work done by it in achieving velocity v. Let a constant force F acts on the body of mass m and move it through a distance s thereby changing its velocity from 0 to v We have v 2 = u 2 +2as =0 2 +2as a = v2 2s Force= F = ma = m v2 2s then Work done = W= F x S = m v2 2s x s = ½ m v2 Kinetic energy is related to momentum as K.E = p2 2m [ K.E = ½ mv 2 = ½ m 1 m mv2 = 2m (mv)2 = p2 2m ( P=mv)] Work - energy theorem. According to work energy theorem, the change in kinetic energy of a particle is equal to the work done on it by the net force. Consider a force F acting on a body of mass m so that its velocity changes from u to v in travelling a distance S. Then work done, W= F.S Now change in KE = ½ m v 2 - ½ m u 2 =½ m (v 2 - u 2 )-----(1) We have v 2 = u 2 +2as v 2 - u 2 =2as eqn(1) ½ m 2as = ma x s = F x s = Work done Work done = change in kinetic energy.

4 Potential Energy:- Potential energy is the energy possessed by a body by virtue of its position or state of strain. For e.g,* a body at a height h above from the ground possess potential energy with respect to earth due to its position. *A compressed spring possess potential energy due to its state of strain. *A stretched bow, water stored in a dam, an elongated spring etc possess potential energy. Gravitational potential energy( U ) of an object of mass m kept at a height h from the ground level. The work done to raise the mass m to a height h against gravitational force = displacement x force in the direction of displacement. U = h x mg = m g h. This much of work done will be stored in it as potential energy. Mechanical Energy: - The sum total of potential and kinetic energy of an object is called mechanical energy. Work - Energy Theorem. According to work energy theorem, the change in kinetic energy of a particle is equal to the work done on it by the net force. Consider a force F acting on a body of mass m so that its velocity changes from u to v in travelling a distance S. Then work done, W = F S. Now change in KE = ½ m (v 2 - u 2 ) But v 2 - u 2 = 2a S Therefore, change in KE = ½ m 2a S = ma S = FS = Work done Thus, workdone = change in kinetic energy. Law of conservation of energy:- According to law of conservation of energy, energy can neither be created nor destroyed. It can be changed from one form to another.

5 1.Proof of law of conservation of energy in case of a freely falling body. Consider an object of mass m kept at a height h at position A. Case (1):- At A. Here energy is entirely potential and is equal to mgh. Mechanical energy at A = PE + KE = mgh +0 = m g h...(1) Case (3):- At B. Now, consider an intermediate position B at a distance x from A during the fall. At B, its potential energy = mg (h - x) = m g h - m g x...(1) K.E = ½ mv 2 From equation of motion, v 2 = u as (u=0, a=g, s=x) here v 2 = 2 g x. ½ mv 2 = ½ m x 2 g x = mgx (2) Mechanical energy at B = PE + KE = mgh - mgx + mgx = m g h...(3) Case (2):- At C. Let the object be released. Just before touching the ground, its energy is entirely kinetic because h = 0, and PE = mgh = 0 Mechanical energy at C = PE + KE = 0 + ½ mv 2 ; where v is the final velocity or velocity at C. Because initial velocity u = 0, a = g and S = h. From equation of motion, v 2 = u as here v 2 = 2 g h. ½ mv 2 = ½ m x 2 g h = mgh (2) which is equal to mechanical energy at B. So the potential energy at position A is completely converted to kinetic energy at position B From (1), (2) and (3) we can see that the mechanical energy of a freely falling body remains constant. 2.Vibration of a simple pendulum:- Pendulum oscillates between A & B. OS (in Figure) is the normal position of the pendulum.

6 *Potential energy is maximum at the extreme ends (A&B) and zero at the mean position (O). *K.E is zero at the extreme ends and maximum at the mean position. Potential Energy of a spring:- Consider a body of mass m attached at the end of a spring suspended from a rigid support. Now let the mass be pulled through a small distance x. Then the spring will try to come back to the initial position by giving an opposite force F. This force is called restoring force. Now restoring force F α x i.e. F = - kx. Here k is called force constant /spring constant of the spring. *The -ve sign shows that the force is opposite to the displacement. Now let the spring be further pulled through a distance dx. Then work done, dw = F dx = k x dx. Therefore the total work done in pulling the spring through a distance x, W = dw = k x dx = ½ k x 2 ; This work done is stored as potential energy in the spring. Therefore the potential energy, U = ½ kx 2 Graphical Method:- Since F α x, a graph drawn connecting F and x is a straight line as shown. The area under the graph gives work done. The area is given by ½ x F {Since it is a triangle and area = ½ base x altitude} But F = k x. Work done, W = ½ x. kx = ½ k x 2 This much work done will be stored in the spring as potential energy. Mass energy equivalence :- Albert Einstein showed that mass and energy are equivalent and they are connected by the relation E = m C 2 ; where E is the energy when m gram of matter is converted and C is the velocity of light. Collissions. A collision is said to have taken place if two moving objects strike each other or come close to each other such that the motion of one of them or both of them changes suddenly. There are two types of collisions. (1) Elastic collision (2) Inelastic collision Elastic collision Elastic collision is one in which both momentum and kinetic energy is conserved.

7 Eg: (1) collision between molecules and atoms (2) collision between subatomic particles. Characteristics of elastic collision (1) Momentum is conserved (2) Total energy is conserved (3) K. E. is conserved (4) Forces involved during collision are conservative forces (5) Mechanical energy is not conserved if it converted into any other form like sound, light heat etc. Inelastic collision Inelastic collision is one in which the momentum is conserved, but KE is not conserved. Example. (1) Mud thrown on a wall (2) Any collision between macroscopic bodies in energy day life. Characteristics of inelastic collision (1) Momentum is conserved (2) Total energy is conserved (3) K.E. is not conserved (4) Forces involved are not conservative (5) Part or whole of the KE is converted into other forms of energy like heat, sound, light etc. Collisions in one dimension If the two objects moves along the same line before and after collision, that is considered as collision in one dimension. Let an object of mass m 1 moving with a velocity u 1, collide with a body of mass m 2 moving with velocity u 2. Momentum is conserved in elastic collision momentum before collision = momentum after collision. m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v (a) m 1 u 1 - m 1 v 1 = m 2 u 2 - m 2 v (1) m 1 (u 1 - v 1 ) = m 2 (v 2 - u 2 ) (2) K.E is conserved in one dimensional collision ½ m 1 u 1 + ½ m 2 u 2 = ½ m 1 v 1 + ½ m 2 v (b) 2 2 m 1 (u 1 - v 1 ) = m 2 (v u 2 ) (3) From eqn.( (3) (2) ) m (u 1 - v 1 ) = m 2 (v u 2 ) m 1 (u 1 - v 1 ) = m 2 (v 2 - u 2 )

8 m 1 (u 1 + v 1 ) (u 1 - v 1 ) = m 2 (v 2 + u 2 ) (v 2 - u 2 ) = m 1 (u 1 - v 1 ) = m 2 (v 2 - u 2 ) (u 1 - u 2 ) = -(v 1 v 2 ) (4) relative velocity before collision = relative velocity after collision. From eqn (4), v 2 = u 1 u 2 + v 1 Substitute the value of v 2 in eqn (a) m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 ( u 1 u 2 + v 1 ) m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 u 1 - m 2 u 2 +m 2 v 1 m 1 u 1 - m 2 u 1 + m 2 u 2 + m 2 u 2 = m 1 v 1 + m 2 v 1 u 1 (m 1 - m 2 ) + 2m 2 u 2 = v 1 (m 1 + m 2 ) v 1 = (m 1- m 2 ) (m 1 + m 2 ) u m 2 (m 1 + m 2 ) u (5) Similarly, v 2 = (m 2- m 1 ) (m 1 + m 2 ) u 2 m (m 1 + m 2 ) u (6) Special cases:- Case(1). Let m 1 = m 2 =m, eqn (5) & (6) becomes v 1 = u 2 and v 2 = u 1 i.e., In one dimensional elastic collision between two bodies of equal mass, the bodies merely exchange their velocities. Case (2):- If the body B is at rest before collision and m 1 m 2, u 2 =0, then from eqn (5), v 1 = (m 1- m 2 ) 2 m 1 (m 1 + m 2 ) & v 2 = (m 1 + m 2 ) u (7) (i).when m 1 = m 2 =m, then eqn (7) v 1 = 0 & v 2 = u 1 ; m 1 comes to rest and m 2 starts moving with initial velocity. Elastic collision in two dimension:- Consider two perfectly elastic balls of masses m 1 & m 2 moving along the same straight line say X-axis with velocities u 1 & u 2 (u 1 > u 2 ).suppose after the collision m 1 & m 2 move off in different directions with velocities v 1 and v 2 making angles θ 1 and θ 2.

9 Applying law of conservation of momentum along X and Y direction, m 1 u 1 + m 2 u 2 = m 1 v 1 Cos θ 1 + m 2 v 2 Cosθ 2 (along X-axis) & 0= m 1 v 1 Sin θ 1 - m 2 v 2 Sin θ 2 (along Y-axis) Since K.E is conserved in collision ½ m 1 u 1 + ½ m 2 u 2 = ½ m 1 v 1 + ½ m 2 v 2