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Page 1 of 23, FUNDAMENTAL OF PHYSICS, for IIT-JEE, NEET, 11th & 12th, Q.1.The sign of work done by a force on a body is important to understand. State carefully if the, following quantities are positive or negative:, (a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket., (b) work done by the gravitational force in the above case,, (c) work done by friction on a body sliding down an inclined plane,, (d) work done by an applied force on a body moving on a rough horizontal plane with uniform, velocity,, (e) work done by the resistive force of air on a vibrating pendulum in bringing it, Solution:, (a) It is clear that the direction of both the force and the displacement are the san, done on it is positive., and thus the work, (b) It can be noted that the displacement of the object is in an upward direction whereas, the force due, to gravity is in a downward direction. Hence, the work done is negative., (c) It can be observed that the direction of motion of the object is, frictional force. So, the work done is negative., oposite to the direction of the, (d) The object which is moving in a rough horizontal plane faces the frictional force which is opposite to, the direction of the motion. To maintain a uniform velocity, a uniform force is applied on the object. So,, the motion of the object and the applied force are in the same direction. Thus, the work done is positive., (e) It is noted that the direction of the bob and the resistive force of air which is acting on it are in, opposite directions. Thus, the work done is negative., Q.2. A body of mass 2 kg initially at, of 7 N on a table with the coefficient of kinetic friction = 0.1. Compute the, (a) work done by the applied force in 10 s,, (b) work done by friction in 10 s,, (c) work done by the net force on the body in 10 s,, (d) change in kinetic energy of the body in 10 s,, moves under the action of an applied horizontal force, Solution:, The mass of the, = 2 kg, Horizontal force applied = 7 N, Coefficient of kinetic friction 0.1, Acceleration produced by the applied force, a1= F/m = 7/2 = 3.5 m/s?, Force of friction, f = uR umg = 0.1 x 2 x 9.8, Retardation produced by friction, a2 = -f/m -196/2 = -0.98, Net acceleration with which the body moves, a = a, + a2 = 3.5 – 9.8 = 2.5, Distance moved by the body in 10 seconds,, S = ut + (1/2)at? = 0 + (1/2) x 2.52 x (10)2= 126 m, for IIT-JEE, NEET, 11th & 12th, R.K.Sharma (8507152734)

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Page 2 of 23, FUNDAMENTAL OF PHYSICS, for IIT-JEE, NEET, 11th & 12th, (a) The time at which work has to be determined is t = 10 s, Work = Force x displacement, = 7 x 126 = 882 J, (b) Work done by the friction in 10 s, W = -f x s = -1.96 x 126, (c) Work done by the net force in 10 s, W = (F - f)s = (7 – 1.96) 126 = 635 J, (d) From v =u + at, V = 0 + 2.52 x 10 = 25.2 m/s, Final Kinetic Energy (1/2) mv2 = (1/2) x 2 x (25.2)2 = 635 J, Initial Kinetic Energy3D (1/2) mu? = 0, Change in Kinetic energy = 635 – 0 = 635 J, The work done by the net force is equal to the final kinetic energy, Q. 3. Given in Figure, are examples of some potential energy functions in one dimension. The, total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify, the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the, minimum total energy the particle must have in each case. Think of simple physical contexts for, which these potential energy shapes are relevant., V3, V., ET, X., V(x)=V(a), a cbd, (a), (b), V, FUNDAMEN OF PHYSICS, V., a, -V., (c), (d), for IIT-JEE, NEET, 11th & 12th, R.K.Sharma (8507152734)

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Page 3 of 23, FUNDAMENTAL OF PHYSICS, for IIT-JEE, NEET, 11th & 12th, Solution:, The total energy is given by E = K.E. + P.E., K.E. = E - P.E., Kinetic energy can never be negative. The particle cannot exist in the region, where K.E. would become, negative., (a) For the region x = 0 and x = a, potential energy is zero. So, kinetic energy is positíve. For, x > a, the, potential energy has a value greater than E. So, kinetic energy becomes zero. Thus the particle will not, exist in the region x > a., The minimum total energy that the particle can have in this case is zero., (b) For the entire x-axis, P.E. > E, the kinetic energy of the object would be negative. Thus the particle, will not exist in this region., (c) Here x = 0 to x = a and x > b, the P.E. is greater than E, so the k, object cannot exist in this region., energy is negative. The, (d) For x = -b/2 to x -a/2 and x = a/2 to x = b/2. Kinetic energy is positive and the P.E. < E. The particle, is present in this region., Q.4. The potential energy function for a particle executing linear simple harmonic motion is, given by V(x) = kx?/2, where k is the force constant of the oscillator. For k = 0.5 N m“, the graph, of V(x) versus x is shown in Fig. 6.12. Show that a particle of total energy 1 J moving under this, potential must 'turn back' when it reaches, V(x)T, X, Solution:, Particle energy E = 1 J, K= 0.5 N m-1, for IIT-JEE, NEET, 11th & 12th, R.K.Sharma (8507152734)

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Page 4 of 23, FUNDAMENTAL OF PHYSICS, for IIT-JEE, NEET, 11th & 12th, K.E = ! mv2, Based on law of conservation of energy:, E = V +K, 1 =, kx? + mv?, Velocity becomes zero when it turns back, 1= kx?, * x0.5x2 = 1, X2 = 4, X= +2, Thus, on reaching x = +2 m, the par, eturns back., Q.5. Answer the following:, (a) The casing of a rock, required for burning obtained? The rocket or the atmosphere?, in flight burns up due to friction. At whose expense is the heat energy, (b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet, due to the sun is not normal to the comet's velocity in general. Yet the work done by the, gravitational force over every complete orbit of the comet is zero. Why?, deaMENTAL OF PHYSICS, (c) An artificial satellite orbiting the earth in a very thin atmosphere loses its energy, gradually due to dissipation against atmospheric resistance, however small. Why, then does its speed increase progressively as it comes closer and closer to the earth?, (d) In Fig. 6.13(i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig., he walks the, same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg, hangs at its other end. In which case is the work done greater?, for IIT-JEE, NEET, 11th & 12th, R.K.Sharma (8507152734)

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Page 5 of 23, FUNDAMENTAL OF PHYSICS, for IIT-JEE, NEET, 11th & 12th, 15 kg, 15 kg, Solution:, (a) When the casing burns up due to the friction, the rocket's mass gets reduced., As per the law of conservation of energy:, Total energy = kinetic energy + potential energy, = mgh +, mv?, There will be a drop in total energy due to the reduction in the mass of the rocket. Hence, the energy, which is needed for the burning, the casing is obtained from the rocket., (b) The force due to gravity is a conservative force. The work done on a closed path by the, conservative force is zero. Hence, for every complete orbit of the comet, the work done by the, gravitational force is zero., (c) The potential energy of the satellite revolving the Earth decreases as it approaches the Earth and, since the system's total energy should remain constant, the kinetic energy increases. Thus, the, satellite's velocity increases. In spite of this, the total energy of the system is reduced by a fraction due, to the atmospheric friction., (d), Scenario I:, m = 20 kg, Displacement of the object, s = 4 m, W = Fs cos e, for IIT-JEE, NEET, 11th & 12th, R.K.Sharma (8507152734)