For JEE Main other Engineering Entrance Exam Preparation, JEE Main Physics Work, Energy And Power<\/strong> Previous Year Questions with Solutions is given below.<\/p>\n Multiple Choice with ONE correct answer<\/strong><\/span><\/p>\n 1.<\/strong> Two masses of 1 g and 4 g are moving with equal kinetic energy. The ratio of the magnitudes of their momenta is [1980-2 marks]\n Given: m\u2081 = 1g, m\u2082 = 4g, KE\u2081 = KE\u2082 2.<\/strong> If a machine is lubricated with oil [1980-2 marks]\n a) the mechanical advantage of the machine increases Lubrication reduces friction \u2192 less energy loss \u2192 increases mechanical efficiency but does not affect mechanical advantage. 3.<\/strong> A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration ac is varying with time \u2018t\u2019 as ac = k\u00b2 r t\u00b2 where \u2018k\u2019 is a constant. The power delivered to the particle by the force acting on it is [1994-1 mark]\n Given: a_c = k\u00b2 r t\u00b2 \u2192 v = k r t 4.<\/strong> A stone tied to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position, and has a speed u. The magnitude of the change in its velocity as it reaches a position where the string is horizontal is [1998-2 marks]\n Velocity at lowest point = u vertically upward 5.<\/strong> A spring of force constant k is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force constant of [1999-2 marks]\n a) (2\/3) k b) (3\/2) k c) 3k d) 6k<\/p>\n Force constant inversely proportional to length. 6.<\/strong> A wind-powered generator converts wind energy into electrical energy. Assume that the generator converts a fixed fraction of the wind energy intercepted by its blades into electrical energy. For wind speed v, the electrical power output will be proportional to [2000]\n a) v b) v\u00b2 c) v\u00b3 d) v\u2074<\/p>\n Wind power is proportional to kinetic energy flux: P \u221d \u00bd \u03c1 A v\u00b3 7.<\/strong> A particle, which is constrained to move along the x-axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as F(x) = -kx + ax\u00b3. Here k and a are positive constants. For x > 0, the functional graphical form of the potential energy U(x) of the particle is [2000]\n Potential energy U(x) = -\u222bF(x) dx = -\u222b(-kx + a x\u00b3) dx = \u222b(kx – a x\u00b3) dx = (k x\u00b2)\/2 – (a x\u2074)\/4 + C 8.<\/strong> An ideal spring with spring constant k is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring is [2002]\n At max extension x: 9.<\/strong> If W\u2081, W\u2082 and W\u2083 represent the work done in moving a particle from A to B along three different paths 1, 2 and 3 respectively (as shown) in the gravitational field of a point mass m, find the correct relation between W\u2081, W\u2082 and W\u2083 [2003]\n a) W\u2081 > W\u2082 > W\u2083 Gravitational field is conservative \u2192 work done independent of path. 10.<\/strong> A particle is acted by a force F = kx, where k is a +ve constant. Its potential energy at x = 0 is zero. Which curve correctly represents the variation of potential energy of the block with respect to x? [2004]\n Since F = kx (positive), potential energy U = -\u222bF dx = – (k x\u00b2)\/2 + C 12.<\/strong> A block of mass 2 kg is free to move along the x-axis. It is at rest and from t = 0 onwards it is subjected to a time-dependent force F(t) in the x direction. The force F(t) varies with t as shown in the figure. The kinetic energy of the block after 4.5 seconds is<\/p>\n a) 4.50 J b) 7.50 J Data insufficient to calculate here (force-time graph missing).<\/p>\n 13.<\/strong> A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time t is proportional to [1984-2 marks]\n Power = force \u00d7 velocity = constant 14.<\/strong> A uniform chain of length L and mass M is lying on a smooth table and one third of its length is hanging vertically down over the edge of the table. If g is acceleration due to gravity, the work required to pull the hanging part onto the table is [1985-2 marks]\n a) MgL b) MgL\/3 Work done = Weight \u00d7 height lifted 15.<\/strong> A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that [1987-2 marks]\n a) its velocity is constant Since force perpendicular to velocity: 17.<\/strong> A small ball starts moving from A over a fixed track as shown in the figure. Surface AB has friction. From A to B the ball rolls without slipping. Surface BC is frictionless. K_A, K_B and K_C are kinetic energies of the ball at A, B and C respectively. Then [2006-5 marks]\n a) h_A > h_C; K_B > K_C Due to friction between A and B, mechanical energy reduces: 18.<\/strong> The speed of the block at point B immediately after it strikes the second incline is<\/p>\n Data incomplete for numeric answer here.<\/p>\n 19.<\/strong> The speed of the block at point C, immediately before it leaves the second incline is<\/p>\n Data incomplete for numeric answer here.<\/p>\n 20.<\/strong> If the collision between the block and the incline is completely elastic, then the vertical (upward) component of the velocity of the block at point B, immediately after it strikes the second incline is<\/p>\n (c) The velocity of the block coming down from the incline AB makes an angle 30\u00b0 with the incline BC. If the block collides with the incline BC elastically, the angle of velocity of the block after collision with the incline shall be 30\u00b0. Hence just after collision with the incline BC the velocity of block shall be horizontal. So immediately after the block strikes the second incline, its vertical component of velocity will be zero.<\/p>\n 22.<\/strong> A pulley arrangement is shown in the figure. The cylinder A has a diameter of 30 cm and the cylinder B has a diameter of 20 cm. The working handle has an arm of 50 cm. The direction of winding of the rope on A is opposite to that on B. Calculate the mechanical advantage of this arrangement. [1980]\n Data insufficient to calculate exact value here.<\/p>\n 23.<\/strong> A body of mass 2kg is being dragged with a uniform velocity of 2m\/s on a rough horizontal plane. The coefficient of friction between the body and the surface is 0.20, J=4.2 joule\/calorie and g = 9.8 m\/s\u00b2. Calculate the amount of heat generated in 5 seconds. [1980-5 marks]\n Friction force = \u03bcmg = 0.20 \u00d7 2 \u00d7 9.8 = 3.92 N 24.<\/strong> In the figures (a) and (b), AC, DG and GF are fixed inclined planes, BC = EF = x and AB = DE = y. Small block of mass is released from the point A. It slides down AC and reaches C with a speed v_c. The same block is released, from rest, from the point D. It slides down DGF and reaches the point F with speed v_F. The coefficients of kinetic friction between the block and both the surfaces AC and DGF are \u03bc. Calculate v_c and v_F. [1980-6 marks]\n Data insufficient for numeric answer here.<\/p>\n 25.<\/strong> A lead bullet just melts when stopped by an obstacle. Assuming that 25 percent of the heat is absorbed by the obstacle, find the velocity of the bullet. Its initial temperature is 27\u00b0C. (Melting point of lead = 327\u00b0 C, specific heat of lead = 0.03 calorie\/g\u00b0C, latent heat of fusion of lead = 6 cal\/g, J = 4.2 joule\/calorie). [1981-3 marks]\n Data insufficient for numeric answer here.<\/p>\n 26.<\/strong> Two blocks A and B are connected to each other by a string and a spring; the string passes over a frictionless pulley as shown in the figure. Block B slides over the horizontal top surface of a stationary block C and the block A slides along the vertical side of C, both with the same uniform speed. The coefficient of friction between the surfaces of blocks is 0.2. Force constant of the spring is 1960 N\/m. If mass of block A is 2kg, calculate the mass of block B and the energy stored in the spring. [1982-5 marks]\n Both A and B move at uniform speed \u2192 acceleration zero. 27.<\/strong> A 0.5 kg block slides from the point A (see fig) on a horizontal track with an initial speed of 3 m\/s towards a weightless horizontal spring of length 1 m and force constant 2 N\/m. The part AB of the track is frictionless and the part BC has coefficients of static and kinetic friction as 0.22 and 0.2 respectively. If the distances AB and BD are 2 m and 2.14 m respectively, find the total distance through which the block moves before it comes to rest completely. (Take g = 10 m\/s\u00b2). [1983-7 marks]\n The part AB is frictionless \u2192 no energy lost. 28.<\/strong> A string, with one end fixed on a rigid wall, passing over a fixed frictionless pulley at a distance of 2 m from the wall, has a point mass M = 2 kg attached to it at a distance of 1 m from the wall. A mass m = 0.5 kg attached at the free end is held at rest so that the string is horizontal between the wall and the pulley and vertical beyond the pulley. What will be the speed with which the mass M will hit the wall when the mass m is released? [1985-6 marks]\n Data insufficient for numeric answer here.<\/p>\n 29.<\/strong> A spherical ball of mass m is kept at the highest point in the space between two fixed, concentric spheres A and B (see figure). The small sphere A<\/p>\n Question incomplete.<\/p>\n 30.<\/strong> A block of mass 0.18 kg is attached to a spring of force-constant 2 N\/m. The coefficient of friction between the block and the floor is 0.1. Initially the block is at rest and the spring is unstretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block in m\/s is V = N\/10. Then N is [2011]\n Data insufficient for numeric answer here.<\/p>\n","protected":false},"excerpt":{"rendered":" JEE Main Physics Work, Energy And Power Previous Year Questions with Solutions For JEE Main other Engineering Entrance Exam Preparation, […]<\/p>\n","protected":false},"author":31,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_yoast_wpseo_focuskw":"JEE Main Physics","_yoast_wpseo_title":"JEE Main Physics Work, Energy And Power Previous Year Questions with Solutions","_yoast_wpseo_metadesc":"Get JEE Main Physics Work, Energy And Power Previous Year Questions with Solutions to infinity learn","custom_permalink":"study-materials\/physics\/jee-main\/work-energy-and-power-previous-year-questions-with-solutions\/"},"categories":[4,21],"tags":[],"table_tags":[],"acf":[],"yoast_head":"\n
\nFrom KE: (1\/2)m\u2081v\u2081\u00b2 = (1\/2)m\u2082v\u2082\u00b2 \u2192 m\u2081v\u2081\u00b2 = m\u2082v\u2082\u00b2
\nMomentum ratio: p\u2081\/p\u2082 = m\u2081v\u2081 \/ m\u2082v\u2082
\nUsing KE relation: v\u2081 = \u221a(m\u2082\/m\u2081) v\u2082 = 2 v\u2082
\nThus, p\u2081\/p\u2082 = (m\u2081 * 2 v\u2082) \/ (m\u2082 v\u2082) = 2(m\u2081\/m\u2082) = 2*(1\/4) = 1\/2
\nAnswer:<\/strong> \u00bd<\/p>\n
\nb) the mechanical efficiency of the machine increases
\nc) both its mechanical advantage and efficiency increase
\nd) its efficiency increases, but its mechanical advantage decreases.<\/p>\n
\nAnswer:<\/strong> (b) the mechanical efficiency of the machine increases.<\/p>\n
\nTangential acceleration a_t = dv\/dt = k r
\nTangential force F_t = m a_t = m k r
\nPower P = F_t \u00d7 v = m k r \u00d7 k r t = m k\u00b2 r\u00b2 t
\nAnswer:<\/strong> P = m k\u00b2 r\u00b2 t<\/p>\n
\nVelocity at horizontal position = u horizontally
\nChange in velocity magnitude = \u221a(u\u00b2 + u\u00b2) = u\u221a2
\nAnswer:<\/strong> u\u221a2<\/p>\n
\nIf total length L, pieces are L\/3 and 2L\/3.
\nLong piece force constant = k \u00d7 (L \/ length of piece) = k \u00d7 (L \/ (2L\/3)) = (3\/2)k
\nAnswer:<\/strong> (b) (3\/2) k<\/p>\n
\nAnswer:<\/strong> (c) v\u00b3<\/p>\n
\nAnswer:<\/strong> U(x) = (k\/2) x\u00b2 – (a\/4) x\u2074 + C<\/p>\n
\nGravitational PE lost = Spring PE gained
\nM g x = (1\/2) k x\u00b2 \u2192 x = (2 M g) \/ k
\nAnswer:<\/strong> x = (2 M g) \/ k<\/p>\n
\nb) W\u2081 = W\u2082 = W\u2083
\nc) W\u2081 < W\u2082 < W\u2083
\nd) W\u2082 > W\u2081 > W\u2083<\/p>\n
\nAnswer:<\/strong> (b) W\u2081 = W\u2082 = W\u2083<\/p>\n
\nU(x) is a downward parabola with maximum at x = 0.
\nAnswer:<\/strong> Parabola opening downwards with maximum at origin.<\/p>\n
\nc) 5.06 J d) 14.06 J<\/p>\n
\nVelocity v \u221d t
\nDistance s = \u222bv dt \u221d \u222b t dt \u221d t\u00b2
\nAnswer:<\/strong> s \u221d t\u00b2<\/p>\n
\nc) MgL\/9 d) MgL\/18<\/p>\n
\nMass of hanging part = M\/3
\nCenter of mass of hanging part = (L\/3)\/2 = L\/6 below table
\nWork = (M\/3) g (L\/6) = MgL\/18
\nAnswer:<\/strong> (d) MgL\/18<\/p>\n
\nb) its acceleration is constant
\nc) its kinetic energy is constant
\nd) it moves in a circular path<\/p>\n
\nVelocity magnitude constant \u2192 kinetic energy constant
\nMotion is circular in plane
\nAnswer:<\/strong> (c) and (d)<\/p>\n
\nb) h_A > h_C; K_B < K_C
\nc) h_A = h_C; K_B = K_C
\nd) h_A < h_C; K_B > K_C<\/p>\n
\nHeight h_A > h_C
\nKinetic energy K_B > K_C (frictionless BC)
\nAnswer:<\/strong> (a)<\/p>\n
\nDistance moved in 5s = velocity \u00d7 time = 2 \u00d7 5 = 10 m
\nWork done = heat generated = force \u00d7 distance = 3.92 \u00d7 10 = 39.2 J<\/p>\n
\nForce on spring (Tension) balances friction.
\nData insufficient for numeric answer here.<\/p>\n
\nThe block compresses spring by x.
\nKinetic energy = Work done against friction + potential energy in spring.
\nData insufficient for numeric answer here.<\/p>\n