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01_Unit_and_Dimension.pdf 02_Motion_in_1-D (1).pdf 04_Motion_in_2-D1.pdf 05_Laws_of_Motion_and_Frictions.pdf 06_Work_power_and_Energy.pdf 09_Magnetism.pdf 10__Electromagnetic_Induction.pdf

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    WORK, ENERGY AND POWER STRAIGHT LINES INTRODUCTION : The term 'work' as understood in everyday life has a different meaning in scientific sense. If a coolie is carrying a load on his head and waiting for the arrival of the train, he is not performing any work in the scientific sense. In the present study, we shall have a look into the scientific aspect of this most commonly used term i.e., work. WORK (a) (b) Example 1 : Solution: (a) (b) Example 2 : Solution: POWER Work in terms of rectangular components Work done by a variable force YB Fxdx + Fydy + Fzdz YB A particle moving in the xy plane undergoes a displacement s = (8.0 i + 6.0 j) m while a constant force F = (4.0 i + 3.0 j) N acts on the particle. (a) Calculate the magnitude of the displacement and that of the force. (b) Calculate the work done by the force. s = + (8.0)2 = 10 m F = + $ — (4.0)2 + (3.0)2 Work done by force, W = F . s = + 3.0 j) • (8.0 i) + 6.0 j) N.m = 32 +0+0+ 18 = 50 N.m = 50J A force F = (4.0 x i + 3.0 yj) N acts on a particle which moves in the x-direction from the origin to x = 5.0 m. Find the work done on the object by the force. Here the work done is only due to x component of force because displacement is along x- axis. Fxdx = 50 J A unit power is the power of an agent which does unit work in unit time. The power of an agent is said to be one watt if it does one joule of work in one second. 1 watt = 1 joule/section = 107 erg/second
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    1 newtonxl metre Also, 1 watt = 1 second [ML2T-2 ] [Work] = [ML2 T-3] Dimensional formula of power : [Power] = [Time] Power has 1 dimension in mass, 2 dimensions in length and — 3 dimensions in time. S.No. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Human Activity Heart beat Sleeping Sitting Riding in a car Walking (4.8 km In-I) Cycling (15 km In-I) Playing Tennis Swimming (breaststroke, 1.6 km h Skating Climbing Stairs (1 16 steps min-I) Cycling (21.3 km In-I) Playing Basketball Tube light Fan Power (W) 1.2 83 120 140 265 410 440 475 535 685 700 800 40 60 UNIT OF WORK / POWER 1 Joule = 107 erg 1 erg = 10-7 joule 1 1.6 x 10 19 J 1 = 3.6 x 106 J 1 joule = 6.25 x 1012 Mev 1 Joule sec-I = 1 watt POWER dw dk = F x V = rate of doing work; Power = p _ dt - dt delivered in projectile motion P is given by P = mg2t + (— mgu Sino ) Where u = velocity of projection, 0 = angle of projection. pys t 10-pcjaÄile av time power 746 watt = 1 hp Example 3 : Solution : What is the power of an engine which can lift 20 metric ton of coal per hour from a 20 metre deep mine? Mass, m = 20 metric ton = 20 x 1000 kg; Distance, S = 20 m; Time t = , 1 = 3600 s Work Power = Time mgxs watt = 1.09 x 103 W 3600
  • 3
    ENERGY Definition: Energy is defined as internal capacity of doing work. When we say that a body has energy we mean that it can do work. Energy appears in many forms such as mechanical, electrical, chemical, thermal (heat), optical (light), acoustical (sound), molecular, atomic, nuclear etc., and can change from one form to the other. In this section, we restrict ourselves to mechanical energy which comprises of two forms: (i) kinetic energy (ii) potential energy (i) Kinetic Energy : Kinetic energy is the internal capacity of doing work of the object by virtue of its motion. K = —m(v.v; ), m = mass, v = velocity Typical kinetic energies (K) -l) S.No. 1 2 3 4 5 6 Object Air molecule Rain drop at terminal speed Stone dropped from 10 m Bullet Running athlete Car Mass (kg) 26 3.5 105 1 5 x 105 70 2000 Speed (m s 500 9 14 200 10 25 2mEK 10 21 103 1.4 X 2 10 3 10 3.5 x 103 105 6.3 x Relation Between Momentum and Kinetic Energy : p (a) (b) Potential Energy : Potential energy is the internal capacity of doing work of a system by virtue of its configuration. É.di = -W Important points for P.E. (i) Potential energy can be defined only for conservative forces. (ii) Potential energy can be positive or negative, depending upon choice of frame of reference. (iii) Potential energy depends on frame of reference but change in potential energy is independent of reference frame. (iv) Potential energy should be considered to be a property of the entire system, rather than assigning it to any specific particle. Types of potential energy Elastic potential energy U - — ky2 where k is force constant and 'y' is the stretch or compression. energy is always positive. Electric potential energy Elastic potential
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    1 qq2 41t€o As charge can be positive or negative, therefore electric potential energy can also be positive or negative. (iii) Gravitational potential energy ml 1772 Which for a body of mass 'm' at height 'h' relative to surface of the earth reduces to U mgh. Gravitational potential energy can be positive or negative. Mechanical Energy Definition : Mechanical energy 'E of an object or a system is defined as the sum of kinetic energy 'K and potential energy 'U, i.e., E K+U (a) (b) (c) (d) Example 4 : Solution: It is a scalar quantity having dimensions [ML 2 T 2] and Sl units joule. It depends on frame of reference. A body can have mechanical energy without having either kinetic energy or potential energy. However, if both kinetic and potential energies are zero, mechanical energy will be zero. The converse may or may not be true, i.e., if E O either both PE and KE are zero or PE may be negative and KEmay be positive such that KE+PE O. As mechanical energy E—K+U, i.e., E —U K. Now as Kis always positive, E U > 0, for existence of a particle in the field, E > U. As mechanical energy E K + U and Kis always positive, so if 'U is positive 'E will be positive. However, if potential energy U is negative, 'E will be positive if K > IUI and E will be negative if UI i.e., mechanical energy of a body or system can be negative, and negative mechanical energy means that potential energy is negative and in magnitude it is more than kinetic energy. Such a state is called bound state, e.g., electron in an atom or a satellite moving around the planet are in bound state. A 2 kg block initially at rest is pulled to the right along a horizontal frictionless surface by a constant force of 12 N, as shown in the figure. Find (a) The speed of the block after it has moved 3.0 m. (b) The acceleration of the block and its final speed using the kinematic equation v; =Vi2 +2as. F = 12 N The normal force balances the weight of the block, and neither of these forces does work since the displacement is horizontal. Since there is no friction, the resultant external force is the 12 N force. The work done by this force is (12 N) (3.0 m) = 36 N.m = 36J Using the work-energy theorem and noting that the initial kinetic energy is zero, we get 1 mvf —0 2 2W 2(36J) 36 m2 / s2 m 210 w = 6 m/s
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    1. 2. 3. 4. (b) Difference between Conservative force and non-conservative force k 5 = 5 m/s a = 6.0 m/s2; 6 m/s. Note that result calculated in two ways are same. Conservative Force Work done by conservative force depends on initial and final positions, not on path followed. Work done by conservative force around a closed loo is zero. Conservative force is related with P.E. as 1. 2. 3. 4. Non-Conservative Force Work done by non-conservative force does depend on the actual path traversed. Work done by non-conservative force around a closed loo is non-zero. Work done by non-conservative force is dw = —dlJ and F(x) dlJ where dx related with heat energy. Friction in liquid conservative force. energy and kinetic dU = change in P.E. and dw is the work done b conservative force Gravitational force/spring force/ electrostatic force are examples of conservative force. or solid are non- Example 5 : Solution: Energy Laws A block of mass m = Ikg is pushed against a spring of spring constant k =100 N/m fixed at one end to a wall. The block can slide on a frictionless table as shown in Figure. The natural length of the spring is Lo = 1m and it is compressed to half its natural length when the block is released. Find the velocity of the block. When the block is released the spring pushes it towards right. The velocity of the block increases till the spring acquires its natural length. Thereafter the block loses contact with the spring and moves with constant velocity. Initially the compression in the spring 2 When the distance of block from the wall becomes x where x < Lo the compression is x). Using the principle of conservation of energy 1 2 2 1 2 2 Solving this, v= k/ m -L When the spring acquires x = G, we have then v = (Lo its 1 m 2 natural length 2. Work Energy Therom : Wnet = work done by all the forces(external, internal, conservatives, non consevatives) - AK-K -K =changein K.E Where Kf is final kinetic energy and Ki is the initial kinetic energy. wext + wnc + wc = AK where wc AU Where W = work done by external forces. ext
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    3. 4. Wnc = work done by non conservative force. Wc = work done by conservative force. AK = Change in K.E AU = change in P.E W ext + Wnc = (Kf + Uf) - (Ki + Ui) = change in potential energy Conservation of mechanical energy (if W ext + Wnc = 0, AME = 0) Ui + ki = Uf + Kf, under conservative force total mechanical energy remains conserved Ui = initial P.E and Uf = Final P.E MOTION IN A VERTICAL CIRCLE (a) (b) (c) Example 6 : Solution: Condition of looping the loop is (u > 5gR) gR,N-O u u — 5gR, N —6mg Condition of leaving the circle is 2gR < u < 5gR u Condition of oscillation is 0 < u < 2gR u A heavy particle hanging from a fixed point by a light inextensible string of length C is projected horizontally with speed (g z) . Find the speed of the particle and the inclination of the string to the vertical at the instant of the motion when the tension in the string is equal to the weight of the particle. Let tension in the string becomes equal to the weight of the particle when particle reaches the point B and deflection of the string from vertical is 0. Resolving mg along the string and perpendicular to the string, we get net radial force on the particle at B i.e. F R = T mg COSO If VB be the speed of the particle at B, then o mgsin0 mgcos0 mg
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    2 mvB From (i) and (ii), we get, 2 mvB T mg cos 0 = Since at B, T = mg 2 mvB — cos 0) = = g C ( 1 — COSO) (iii) (iv) Applying conservation of mechanical energy of the particle at point A and B, we have 1 2 mg! (1 —cos0) + —mv{ ; —mv 2 where and gc(l —cos0) gc = 2gC(1 - cos0) + gC (1 - cos0) 2 COS 0 = 3 Putting the value of cos0 in equation (iv), we get : v (v) 3


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