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CBSE Class 11 NCERT Exemplar Physics Work, Energy and Power
Work, Energy and Power

A balloon filled with helium rises against gravity increasing its potential energy. The speed of the balloon also increases as it rises. How do you reconcile this with the law of conservation of mechanical energy? You can neglect the viscous drag of air and assume that the density of air is constant.

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The net buoyant force Equals vpg when the dragging viscous force of the air on the balloon is ignored. Where v is the volume of air that has been displaced. The upward net density is denoted by p....

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A rocket accelerates straight up by ejecting gas downwards. In a small time interval ∆t, it ejects a gas of mass ∆m at a relative speed u. Calculate KE of the entire system at t + ∆t and t and show that the device that ejects gas does work = (1/2) ∆m u2 in this time interval.

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M is the rocket's mass at any given moment t. The rocket's velocity is v. The mass of the gas expelled during the time interval t is m. As a result,  K = 1/2 u2∆m

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A curved surface as shown in the figure. The portion BCD is free of friction. There are three spherical balls of identical radii and masses. Balls are released from one by one from A which is at a slightly greater height than C. with the surface AB, ball 1 has large enough friction to cause rolling down without slipping; ball 2 has a small friction and ball 3 has a negligible friction. a) for which balls is total mechanical energy conserved? b) which ball can reach D? c) for balls which do not reach D, which of the balls can reach back A?

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a) For ball 1 the total mechanical energy is conserved b) Ball 1 reaches D c) Ball 3 reaches back A

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A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of 30o by a force of 10 N parallel to the inclined surface. The coefficient of friction between the block and the incline is 0.1. If the block is pushed up by 10 m along the incline, calculate a) work done against gravity b) work done against the force of friction c) increase in potential energy d) increase in kinetic energy e) work done by an applied force

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a) Work against gravity equals mgh 5 m= h 50 J WD against gravity b) The work done against the friction force is fs = 53 J. d) WD against gravity = 50 J increase in PE d) The system's increase in KE...

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On complete combustion, a litre of petrol gives off heat equivalent to 3 × 107 J. In a test drive a car weighing 1200 kg, including the mass of driver, runs 15 km per litre while moving with a uniform speed on a surface and air to be uniform, calculate the force of friction acting on the car during the test drive, if the efficiency of the car engine were 0.5.

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car engine Efficiency = 0.5 Energy given by the car with 1 litre of petrol = 1.5 × 107 WD = 1.5 × 107 f = 103 N

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An adult weighing 600 N raises the centre of gravity of his body by 0.25 m while taking each step of 1 m length in jogging. If he jogs for 6 km, calculate the energy utilized by him in jogging assuming that there is no energy loss due to friction of ground and air. Assuming that the body of the adult is capable of converting 10% of energy intake in the form of food, calculate the energy equivalents of food that would be required to compensate energy utilized for jogging.

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The energy used up is given as = mgh mg = 600 N h = 0.25m No.of steps in 6 km = 6000 steps Energy used in 6000 m = (6000)(600)(0.25)J Energy utilized in jogging = 9 × 104 J

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An engine is attached to a wagon through a shock absorber of length 1.5 m. The system with a total mass of 50,000 kg is moving with a speed of 36 km/h when the brakes are applied to bring it to rest. In the process of the system being brought to rest, the spring of the shock absorber gets compressed by 1.0 m. If 90% of the energy of the wagon is lost due to friction, calculate the spring constant.

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KE = 1/2 mv2 m = 50000 kg v = 10 m/s KE = 2500000J KE of spring = 10% of the KE wagon K = 5 × 105 N/m

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The bob A of a pendulum released from horizontal to the vertical hits another bob B of the same mass at rest on a table as shown in the figure. If the length of the pendulum is 1 m, calculate a) the height to which bob A will rise after collision b) the speed with which bob B starts moving. Neglect the size of the bobs and assume the collision to be elastic.

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a) After the impact, bob A does not rise much because the PE of bob A is converted to KE and the momentum is transferred to bob B. (B) The speed of bob B is calculated as the sum of bob A's KE and...

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Consider a one-dimensional motion of a particle with total energy E. There are four regions A, B, C, and D in which the relation between potential energy V, kinetic energy (K) and total energy is as given below: Region A: V > E Region B: V < E Region C: K > E Region D: V > K State with reason in each case whether a particle can be found in the given region or not.

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E = V + K and V > E for area A, implying that the KE is negative and therefore this is not feasible. K = E – V and V E for area B, implying that both energies are larger than zero. V = E – K and...

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A ball of mass m, moving with a speed 2v0 collides inelastically with an identical ball at rest. Show that a) for a head-on collision, both the balls move forward b) for a general collision, the angle between the two velocities of scattered balls is less than 90o.

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a) Let v1 and v2 be the velocities of the two balls after the collision. According to the law of conservation of momentum, mv0 = mv1 + mv2 v2 = v1 + 2ev0 e < 1 b) Using the law of conservation of...

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Two blocks M1 and M2 having equal mass are free to move on a horizontal frictionless surface. M2 is attached to a massless spring as shown in the figure. Initially, M2 is at rest and M1 is moving toward M2 with speed v and collides head-on with M2.;a) while spring is fully compressed all the KE of M1 is stored as PE of spring b) while spring is fully compressed the system momentum is not conserved, though final momentum is equal to the initial momentum c) if spring is massless, the final state of the M1 is the state of rest d) if the surface on which blocks are moving has friction, then a collision cannot be elastic

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c) if spring is massless, the final state of the M1 is a state of rest d) if the surface on which blocks are moving has friction, then a collision cannot be elastic

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A bullet of mass m fired at 30o to the horizontal leaves the barrel of the gun with a velocity v. The bullet hits a soft target at a height h above the ground while it is moving downward and emerges out with half the kinetic energy it had before hitting the target. Which of the following statements are correct in respect of bullet after it emerges out of the target? a) the velocity of the bullet will be reduced to half its initial value b) the velocity of the bullet will be more than half of its earlier velocity c) the bullet will continue to move along the same parabolic path d) the bullet will move in a different parabolic path e) the bullet will fall vertically downward after hitting the target f) the internal energy of the particles of the target will increase

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b) the velocity of the bullet will be more than half of its earlier velocity d) the bullet will move in a different parabolic path f) the internal energy of the particles of the target will...

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A man, of mass m, standing at the bottom of the staircase, of height L, climbs it and stands at its top. a) work done by all forces on man is zero b) work done by all the force on man is zero c) work done by the gravitational force on man is mgL d) the reaction force from a step does not do work because the point of application of the force does not move while the force exists

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b) work done by all the force on man is zero d) the reaction force from a step does not do work because the point of application of the force does not move while the force exists

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A cricket ball of mass 150 g moving with a speed of 126 km/h hits at the middle of the bat, held firmly at its position by the batsman. The ball moves straight back to the bowler after hitting the bat. Assuming that collision between ball and bat is completely elastic and the two remain in contact for 0.001 sec, the force that the batsman had to apply to hold the bat firmly at its place would be a) 10.5 N b) 21 N c) 1.05 × ${10}^{4}$ N d) $2.1 \times {10}^{4}$N

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c) 1.05 × 104 N

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Two inclined frictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track as shown in the figure. Which of the following statement is correct?a) both the stones reach the bottom at the same time but not with the same speed b) both stone reach the bottom with the same speed and stone I reaches the bottom earlier than stone II c) both the stones reach the bottom with the same speed and stone II reaches the bottom earlier than stone I d) both the stones reach the bottom at different times and with different speeds

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c) both the stones reach the bottom with the same speed and stone II reaches the bottom earlier than stone I

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A proton is kept at rest. A positively charged particle is released from rest at a distance d in its field. Consider two experiments; one in which the charged particles is also a proton and in another, a positron. In the same time t, the work done on the two moving charged particles is a) the same as the same force law is involved in the two experiments b) less for the case of a positron, as the positron moves away more rapidly and the force on it weakens c) more for the case of a positron, as the positron moves away from a larger distance d) same as the work done by charged particle on the stationary proton

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c) more for the case of a positron, as the positron moves away from a larger distance

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An electron and a proton are moving under the influence of mutual forces. In calculating the change in the kinetic energy of the system during motion, one ignores the magnetic force of one on another. This is because, a) the two magnetic forces are equal and opposite, so they produce no net effect b) the magnetic forces do no work on each particle c) the magnetic forces do equal and opposite work on each particle d) the magnetic forces are necessarily negligible

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b) the magnetic forces do no work on each particle

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