work, energy and power

Scalar Product Scalar product of two vectors and is given by It is also known as dot product. The result of the scalar product of two vectors is a scalar quantity. When two vectors are parallel, θ = 0°, cos 0° = 1 For unit vectors, Similarly,  When two vectors are perpendicular, θ = 90°, cos 90° = 0It means the dot product of two perpendicular vectors is zero. For unit vectors, ORSimilarly,  Properties of Scalar Product of two vectors Scalar product of two vectors is commutative, i.e., Scalar product is distributive, i.e., Scalar product of a vector with itself gives the square of its magnitude, i.e.,  Dot Product in Cartesian Coordinates Let  Work and Kinetic Energy : The Work Energy Theorem We know that according to the third equation of motion,Multiplying both sides by m/2, we obtain∴Where,= Final kinetic energy= Initial kinetic energyW = FS = Work doneEquation (i) is a special case of work energy (WE) theorem. The change in kinetic energy of a particle is equal to the work done on it by the net force. Work Work is said to be done when the point of application of the forces moves in the direction of the force. If a constant force  is applied on a body and the body has a displacement in the direction of the force as shown in fig, then the work done on the body by the force is given by, When the displacement is not in the direction of as shown in figure, In such a case, we find the work done by resolving into two rectangular components.(i) in the direction of  (where Fx = F cos θ)(ii)  perpendicular to  (where Fy = F sin θ)The work is done along the component  only.Thus,∴ Kinetic Energy The kinetic energy of a mass m having velocity v is, K= It is a scalar quantity.   Work, Energy and Variable Force Work Done by a Variable Force The above given figure shows a variable force. The entire path ABCD is broken into infinitesimally small displacements One such small displacement is from P to Q. Let PQ = dx Small amount of work done from P to Q is, dW = F × dxAlong the small displacement dx, force is constant in magnitude and direction.dW = (PR) × (PQ)dW = Area of the strip PQRS Total work done in moving the body from A to B is, If the displacements are allowed to approach zero, then ∴W = Area of ABCDA Work−Energy Theorem for a Variable Force Suppose, m = Mass of a body u = Initial velocity of the body v = Final velocity of the body a = Acceleration ki = Initial kinetic energy of the body kf = Final kinetic energy of the body The rate of change of kinetic energy is, dk = Fdx On integrating from the initial position (xA) to the final position (xB), we have [From equation (i)] ∴Work done on the body = Increase in K.E. of the body Potential Energy and Conservation of Mechanical Energy Potential energy is the energy possessed by the body by virtue of its position.Two important types of potential energy are Gravitational potential energy Elastic potential energy Gravitational potential energy: It is the energy possessed by the body by virtue of its position above the surface of the earth.Let,m = Mass of a bodyg = Acceleration due to gravity h = Height through which the body is raised.The force applied to just overcome gravitational attraction isF = mgWork done = Force × DistanceORW = (F) × h = mghThis work gets stored as potential energy.∴Gravitational P.E. = V (h) = mghIf h is taken as a variable, thenWhere, F is the gravitational force on the bodyThe negative sign indicates that gravitational force is acting downwards.Thus, Mathematically, the potential energy, V (x) is defined if the force F (x) can be written asThe above equation shows that the work done by a conservative force like gravity in taking the body from initial position, (x1) to final position, (x2) is equal to the difference between the initial and final P.E. of the body. The Conservation of Mechanical Energy Total mechanical energy of a system is always conserved.Total mechanical energy = Potential energy (V) + Kinetic energy (K) Mechanical energy of a system is conserved if the forces doing work on it are conservative. Let us consider a body undergoing a small displacement, Δx under the action of a conservative force, F. According to work energy theorem,Change in K.E = Work doneΔk = F(x) Δx … (i)If the force is conservative, the P.E. function, V(x) can be defined such that−ΔV = F (x) ΔxΔV = − F (x) Δx … (ii)Adding equations (i) and (ii)Δk + ΔV = 0Δ (k + V) = 0, which means(k + V) = Constant. Example of the law of conservation of mechanical energy:Let m be the mass of the body held at A, at a height, h above the ground.At point AThe body is at rest at A.K.E. of the body = 0P.E. of the body = mghT.E. of the body = K.E + P.E. = 0 + mghE1 = mgh … (i)At point CLet the body be allowed to fall freely under gravity so as to strike the ground at C with a velocity, v.From v2 − u2 = 2as ORv2 − 0 = 2 (g) hORv2 = 2gh … (ii)K.E. of the body = mv2 = m (2gh) = mgh …[From equation (ii)]P.E. of the body = mgh = mg (0) = 0Total energy of the body = K.E. + P.E.E2 = mgh + 0 = mgh … (iii)At point BIn free fall, suppose the body crosses point B with a velocity, v1, where AB = xv2 − u2 = 2asORK.E. of the body Height of the body at B = CB = (h − x)∴P.E. of the body at B = mg (h − x)Total energy of the body at B = K.E. + P.E.E3 = mgx + mg (h − x)= mgx + mgh − mgxORE3 = mgh … (iv)From equation (i), (iii) and (iv), we find that E1 = E2 = E3 = mghi.e., the total energy of the body during free fall remains constant at all positions The Potential Energy of a Spring Consider a light and perfectly elastic spring fixed at one end of a rigid support at point O and the other end attached to a block of mass ‘m’. When the block is pulled from its equilibrium position (C) to point A, the restoring force is set up in the spring due to elasticity. The work done in stretching the spring from C to A is stored in the system in the form of potential energy of the block. Let us calculate the P.E of mass m, when the spring is pulled from the mean position C up to a point P, such that CP = x The restoring force set up in the string is given by, F ∝ x F = − kx Where, k is the constant of proportionality known as force constant or spring constant Suppose that the block is further displaced through an infinitesimally small distance, PQ = dx Small work done in increasing the length of the spring by dx is dW = Fdx = kxdx [Considering magnitude only] The work done in increasing the length of the spring by an amount x can be calculated by integrating the above limits x = 0 to x = x i.e, This work done is stored in the system as its potential energy at point P. The potential energy of the system, when the block is pulled up to point A, can be obtained by setting x = r ∴P.E. of the system at point A  If we plot the P.E and K.E against the displacement x, then the graph will be as depicted by the two dotted curves: The Law of Conservation Of Energy and Power Different Forms of Energy Internal energy − The sum of kinetic and potential energies of all the molecules constituting the body is called internal energy. Heat energy − A body possesses heat energy due to the disorderly motion of its molecules. Chemical energy − A body possesses chemical energy because of chemical bonding of its atoms.Exothermic reaction − A chemical reaction in which energy is releasedEndothermic reaction − A chemical reaction in which energy is absorbed Electrical energy − Work has to be done in order to move an electric charge from one point to another in an electric field. This work done appears as the electrical energy of the system. Nuclear energy − When a heavy nucleus (such as U − 235) breaks up into lighter nuclei on being bombarded by a slow neutron, a tremendous amount of energy is released. This energy is known as nuclear energy. Principle of Conservation Of Energy It states that energy can be neither created nor destroyed. It can only be converted from one form to another. Power The rate of doing work is called power. The average power is given by, Where, W is work performed by the agent in time‘t’ Instantaneous power − Limiting value of the average power of an agent in a small time interval, when the time interval approaches zero If ΔW is work done in a small interval Δt, then instantaneous power is defined as Then, Again,, the instantaneous velocity of the particle Therefore, If θ is angle between and, then P = Fv cos θ If θ = 0°, then P = Fv Unit of power − Watt The bigger unit of power is kilowatt (KW). 1 KW = 103 W Collisions Collision between two particles is defined as mutual interaction of the particles for a short interval of time as a result of which the energy and momentum of the interacting particles change. Types of Collision Elastic collision − Those collisions in which both momentum and kinetic energy of the system are conserved Inelastic collision − Those collisions in which momentum of the system is conserved, but kinetic energy is not conserved Elastic Collision in One Dimension Consider that two perfectly elastic bodies A and B of masses M1 and M2 moving with initial velocities u1 and u2 undergo head on collision and continue moving along the same straight line with final velocities v1 and v2. As in an elastic collision, momentum is conserved. ∴ M1u1 + M2u2 = M1v1 + M2v2 … (i) Since kinetic energy is also conserved in an elastic collision, we obtain From equation (i), we obtain M1 (u1 − v1) = M2 (v2 − u2) … (iii) From equation (ii), we obtain Dividing equation (iv) by (iii), we obtain ∴u1 − u2 = v2 − v1 … (v) From equation (v), it follows that in one-dimensional elastic collision, the relative velocity of approach (u1 − u2) before collision is equal to the relative velocity of separation (v2 − v1) after collision. Calculation of velocities after collision: Let us first find the velocity of body A after collision. From equation (v), we have v2 = u1 − u2 + v1 Substituting for v2 in equation (i), we obtain Again from equation (v), we have v1 = v2 − u1 + u2 Substituting for v1 in equation (i), we obtain Special Cases When the two bodies are of equal masses i.e.,M1 = M2 = M (say)From equation (vi), we haveAlso from equation (vii), we have When the target body (B) is at rest:In this case, u2 = 0Substituting u2 = 0 in equations (vi) and (vii), we obtainWhen M2 ≥ M1, in equation (viii) and (ix), M1 can be neglected in comparison to M2 i.e., M1 −M2 ≈ M2 and M1 + M2 ≈ − M2. Therefore, we have Elastic Collision in Two Dimensions Suppose m1, m2 are the masses of two bodies A and B moving initially along X-axis with velocitiesu1 and u2 respectively. When u1 > u2, the two bodies collide. After collision, let body A move with a velocity v1 at an angleθ with X-axis. Let body B move with a velocity v2 at an angle Φwith X-axis. As the collision is elastic, K.E. is conserved. ∴ Total K.E. after collision = Total K.E. before collision In elastic collision, linear momentum is also conserved (along X-axis). ∴ Total linear momentum after collision = Total linear momentum before collision Along Y-axis, linear momentum before collision is zero (as both the bodies are moving along X-axis). After collision, total linear momentum along Y-axis is (m1v1sinθ − m2v2sinΦ). From equations (ii), (iii), and (iv), we have to calculate 4 variables v1, v2, θ, and Φ, which is not possible. We have to measure any one parameter experimentally.