Work, Power and Energy PDF

WORK, ENERGY AND UNIT-IV POWER CHAPTER 5 WORK, ENERGY AND POWE...

WORK, ENERGY AND UNIT-IV POWER CHAPTER 5 WORK, ENERGY AND POWER Work done by a constant force and a variable force; kinetic energy, work-energy theorem, Syllabus Power; Notion of potential energy, potential energy of a spring, conservative forces; non- conservative forces; motion in a vertical circle; elastic and inelastic collisions in one and two dimensions. TOPIC - 1 Work and Power.... TOPIC - 2 Topic-1 Work and Power Energy and Collision.... Revision Notes  Work : Work is done when the body is displaced actually through some distance in the direction of applied force. → → W = F⋅ s W = Fs cos q l Work done is a scalar quantity. However, work done is positive when q lies between 0 (zero) and p/2. Work done is negative when q lies between p/2 and 3p/2. l S. I. unit of work is joule (J) and the C.G.S unit of work is erg, where 1 joule = 107 erg. l Work done by a body does not depend on the time taken to complete the work.  Internal work or zero work: The work in which muscles are strained, but work done is not useful, is called internal work. For example, when a person carrying load keeps on standing at the same place, work done is zero, but he gets tired on account of internal work. Dimensions : [ML2 T–2] Power : Power of a body is defined as the time rate of doing work by the body. Thus, in power, time taken by the body to complete the work is significant. Work done Power = Time taken → → P = F ⋅= v Fv cos θ → → l Here, q is the angle between force F and velocity v of the body. l Unit : 1 W = 1 Js–1 l Dimensions : [ML2T–3] l Power is a scalar quantity.  WORK, ENERGY AND POWER Key Words  Conservative force is a force if work done by or against the force in moving a body depends only on the initial & final positions of the body and not on the nature of path followed between initial and final positions, e.g., gravitational force, electrostatic force between two electric charges, all central forces, etc.  Non-conservative force is a force if work done by or against the force in moving a body from one position to another depends on the path followed between these two positions. e.g., frictional forces, elastic forces, etc. Key Formulae  Work= Force × Displacement in the direction of force → →  W = F. s = Fs cos q  Maximum work When cos q = 0° W = Fs  Minimum Work when q° = 90° Then W = Fs × cos 90° = 0  Work done by variable force : xB   W =  F.d x xA Work done  Power = Time Taken → →  P = F⋅ v  P = Fv cos q. Mnemonics Concept: Positive work, negative work n - Negative Fernandez d'souza Ordered d' donut W With If Force and displacement are in opposite noodles directions work done is negative. Fernandez d' d'souza was Served donut WWith F - Force Pizza. d - Displacement Interpretation: s - Same F - Force d - direction d - Displacement w - Work done o - opposite p - positive d - direction If Force and displacement are in same w - Work done directions work done is positive Very Short Answer Type Questions (1 mark each) Q. 1. What is the work done by Earth’s gravitational Ans. Negative work means that force (or its component) force in keeping the Moon in its orbit in a is opposite to displacement. 1 complete revolution ? U (DDE) Q. 4. A spring is stretched. Is the work done by the Ans. Zero, because gravitational force is a conservative stretching force positive or negative ? A (DDE) force and act perpendicular to direction of motion Ans. Positive because the force and the displacement of Moon. 1 are in the same direction. 1 Q. 2. What do you mean by positive work? U (DDE) Q. 5. Does the work done in raising a load on to a Ans. Positive work means that force (or its component) platform depend upon how fast is it raised ? A is parallel to displacement. 1 Ans. No, because the work done is independent of Q. 3. What do you mean by negative work? U (DDE) time. 1  CBSE Question Bank Chapterwise & Topicwise, PHYSICS, Class-XI Short Answer Type Questions-I (2 marks each) Q. 1. What is conservative force ? R  ∆W   [MSE Chandigarh 2008] P = lim   ∆ t → 0  ∆t  Ans. Refer to Know the Terms. 2 → Q. 2. What is instantaneous power ? R → dr → → F = F⋅ v Ans. Instantaneous power is defined as the limiting dt value of the average power when time tends to → zero. where, v is instantaneous velocity. 1 → → dW F⋅ dr In general = = 1 dt dt Short Answer Type Questions-II (3 marks each) Q. 1. Give conditions for no work. R ∆W Pav =...(i) ½ Ans. In physics no work is said to be done, if ∆t (a) The applied force (F) is zero. A body moving with When P be the instantaneous power, then by def. uniform velocity on a smooth surface has some P = Lt Pav displacement but no external force so in this case ∆t → 0 work done is zero. 1 ∆W (b) The displacement (S) is zero. A labourer standing = Lt ½ ∆t → 0 ∆t with a load on his head does no work. ½ (c) The angle between force and displacement (θ) is dW d or P= = (W)...(ii) π/2 rad or 90°. Then cos θ = cos 90° = 0. Thus, dt dt work done is also zero. In circular motion, → → Now W = F ⋅ s ...(iii) ½ instantaneous work done is always zero because of this reason. 1 where, F = constant force producing a displacement (d) The change in kinetic energy (∆KE) is zero. ½ ∴ From equations (ii) and (iii), Q. 2. Prove that instantaneous power is given by the → → d →→ dot product of force and velocity, i.e., P= F ⋅ v  U P= ( ) dt F ⋅ s Ans. Suppose, ∆W be the amount of work done in a → small time interval ∆t, when Pav be the average ds → → → = F⋅ =F⋅ v 1 power, then ½ dt Topic-2 Energy & Collision Revision Notes  Energy : Energy of a body is defined as the capacity of the body to do the work. Energy is a scalar quantity. It has the same units and dimensions as those of work. Some practical units of energy and their relation with S.I. unit of energy (joule) are :   (i) 1 calorie = 4.2 J (ii) 1 kiloWatt hour (kWh) = 3.6 × 106 J (iii) 1 electron volt (1 eV) = 1.6 × 10–19 J  Work-Energy Theorem : According to this principle, work done by net force in displacing a body is equal to change in kinetic energy of the body and i.e., W = Kf – Ki Kf = final K.E. Ki = initial K.E.  Collisions : When a body strikes against another body such that there is exchange of energy and linear momentum then the two are said to collide. Collisions are of two types :  WORK, ENERGY AND POWER (i)  Perfectly elastic collision is that in which there is no change in kinetic energy of the system, i.e.,  Total K.E. before collision = Total K.E. after collision.  e.g., collisions between atomic and subatomic particles are perfectly elastic collisions. (ii)  Perfectly inelastic collision is that in which K.E. is not conserved. Here, the bodies stick together after impact. Linear momentum is conserved in every collision elastic as well as inelastic, further total energy is also conserved in all such collisions. Kinetic energy alone is not conserved in inelastic collisions. Key Words  Kinetic Energy is the energy possessed by the body by virtue of its motion. It is always positive.  Potential Energy is the energy possessed by the body by virtue of its position. It can be both negative as well as positive.  Gravitational Potential Energy is the energy possessed by the body by virtue of its position with respect to center of Earth or other body.  Potential Energy of spring is the energy associated with the state of compression or expansion of an elastic spring.  Internal Energy is the energy possessed by the body by virtue of particular configuration of its molecules.  Coefficient of Restitution or Coefficient of Resilience is the ratio of relative velocity of separation after collision to the relative velocity of approach before collision. It is denoted by ‘e’. Key Formulae 1  Kinetic Energy (K.E.) = mv2 , 2 where, m = mass, v = velocity of particles  Potential Energy (P.E.) = mgh Velocity (v) = 2gh   Force in Spring (F) = -kx, where, k = spring constant, x = compression.  Mass Energy Equivalence : E = mc2 where, m = mass that disappears E = energy that appears c = velocity of light Relative velocity of separation (after collision)  Coefficient of Resilience (e) = Relative velocity of approach (before collision) v2 –v1 e= u2 –u1 for perfectly elastic collision, e =1 for perfectly inelastic collision, e = 0  Elastic collision in 1- Dimension :  m1 – m2  2m2u2 v1 =   u1 +  m1 + m2  m1 + m2 2m1u1  m2 – m1  v2 = +  u2 m1 + m2  m1 + m2   Inelastic Collision in 1–Dimension : m1u1 v= if u2 = 0 m1 + m2 m1m2u12 Loss in K.E. = 2( m1 + m2 ) CBSE Question Bank Chapterwise & Topicwise, PHYSICS, Class-XI Very Short Answer Type Questions (1 mark each) Q. 1. What is mass energy equivalence? R (DDE) Q. 3. What type of energy is stored in the spring of watch? U Ans. The relation between mass of a particle m and its Ans. Potential energy. 1 equivalent energy (E) is given by E = mc2 where, Q. 4. When an arrow is shot, where from the arrow c = velocity of light in vacuum. 1 acquires its K.E. ? A Q. 2. What is the source of the kinetic energy of the Ans. It is the potential energy of the bend bow which is converted into K.E. of arrow. 1 falling rain drops ? R Q. 5. Can P.E. of an object be negative ?  A Ans. It is the gravitational potential energy which is Ans. Yes, it can be negative when forces involved are being converted into kinetic energy. 1 attractive. 1 Short Answer Type Questions-I (2 marks each) Q. 1. What is the relation between linear momentum Ans. When body (B2) is at rest, we get and KE? U (DDE) m – m2 Ans. We know that, KE of a particle, v1 = 1 u m1 + m2 1 1 K = mv 2 2m1 2 and v2 = u 1 where m is the mass of particle and v is the m1m2 1 velocity Putting m1 = m2 = m, we get v1 = 0 and v2 = u1, 1 mv 2×m 1 i.e., body B1 comes to rest whereas body B2 moves K= 2 m with the velocity of body B1. 1 Q. 4. Is kinetic energy a scalar or a vector ? Give its 1 ( mv ) 2 K= S.I. unit and dimensional formula. U 2 m Ans. Kinetic energy is a scalar. ½ p2 K= {p=mv} S.I. unit of kinetic energy is joule (J) ½ 2m Dimensional formula of kinetic energy is [ML2T–2]. 1  1 ∴p = 2mK Q. 5. A bullet weighing 10 g is fired with a velocity of Q. 2. Give conditions for elastic collision. R 800 ms–1. After passing through a mud wall 1 m Ans. In an elastic collision, thick, its velocity decreases to 100 ms–1. Find the (a) Total momentum is conserved, i.e., total final average resistance offered by the mud wall. momentum is equal to the total initial momentum. A (b) Total mechanical energy is conserved, i.e., total 1 1 Ans. Using F·s = mv2 – mu2 ½ final energy is equal to the total initial energy. 2 2 (c) Total kinetic energy is conserved, i.e., total final 10 where m = 10 g = kg kinetic energy is equal to the total initial kinetic 1000 energy. = 0.01 kg (d) All the forces are of conservative nature, i.e., work and v = 100 ms–1, u = 800 ms–1. ½ done does not depend upon the actual paths. and s = 1 m, (½ mark each) 1 we get F= × 0.01 × (1002 – 8002) Q. 3. Body B1 collides with body B2 of same mass but 2 at rest, what will happen to them if collision is = – 3150 N. 1 elastic. A Short Answer Type Questions-II (3 marks each) Q. 1. State and derive work- energy relationship. Ans. It states that change in kinetic energy of a body is  U [KVS 2013] equal to work done and vice versa. Let a constant → OR force F be applied to a body moving with initial Derive a relationship between kinetic energy → → velocity u , so that its velocity becomes v along and work. [KVS 2008] the direction of force when s is its displacement.  WORK, ENERGY AND POWER Using Newton’s second law of motion we get Ans. Using the relation for kinetic energy, magnitude of force F = ma and from equation 1 of motion, we get v2 – u2 = 2as, where a is the K.E. = mv2 2 acceleration of the body. 1 we get rate of change of K.E. with respect to time as Multiplying both sides by m/2, we get d d 1 2 1 1 (K.E.) = mv  mv2 – mu2= mas dt dt  2  2 2 1 dv 1 1 = ×m×2×v× i.e., mv2 – mu2 = Fs = W 2 dt 2 2 dv = mv i.e., K.E.(f) – K.E.(i) = W 1 dt where K.E.(f) is final kinetic energy and K.E.(i) is mdv initial kinetic energy. But = ma = F 1 dt Thus work done on a body by a net force is equal to the change in kinetic energy of the body. 1 where a is acceleration and F is force. d dx Q. 2. A block initially at rest breaks into two parts of ∴ K.E. = Fv = F masses in the ratio 2 : 3. The velocity of smaller dt dt part is (8i +6 j ) m/s. Find the velocity of bigger or d(K.E.) = Fdx 1 part. A [NCT 2011] Integrating between the initial and final energies, Ans. Let mass of the block = m i.e., K.E.i. and K.E.f and also position, i.e., xi and xf 2 3 After breaking, m1 = m and m2 = m respectively, we get 5 5 K. E. f x → → ∫K.E. d( K.E.) = ∫ xi f Fdx Linear momentum = m1 v1 + m2 v2 ½ i ∴ K.E.(f) – K.E.(i) = W According to law of conservation of momentum Pf = Pi The work energy theorem is thus verified for variable force. 1 → → or m1 v1 + m2 v2 = 0 1 → v1 = velocity of smaller part, Commonly Made Error → v Students commit error while finding the 2 = velocity of bigger part derivatives and the integrals of functions to → 2 3 or m (8i + 6 j ) + m ( v2 ) = 0 ½ prove the required result. 5 5 → 3 1 or m v2 = – m (16i + 12 j ) 5 5 Answering Tip →  16  v2 = –  i + 4 j  1 Students should learn about the basics of  3  derivatives and integrals and their use to get Q. 3. Prove work energy theorem for a variable force. the desired result.  U [NCT 2008, 09; MSE Chandigarh 2008, 09] Long Answer Type Questions (5 marks each) Q. 1. What are the various types of equilibrium? U (iv) Example: A marble placed at the bottom of a Ans. Different types of equilibrium are— hemispherical bowl. Stable : Unstable : (i)  When a particle is displaced slightly from a (i) When a particle is displaced slightly from a position, then a force acting on it brings it back position, then a force acting on it tries to displace to the initial position, It is said to be in stable equilibrium position. the particle further away from the equilibrium position, it is said to be in unstable equilibrium (ii) Potential energy is minimum. position. d 2U (iii) = positive (ii) Potential energy is maximum dx 2  CBSE Question Bank Chapterwise & Topicwise, PHYSICS, Class-XI d 2U collision, the two bodies move together with a (iii) = negative common velocity v. As the total linear momentum dx 2 of the system cannot change, therefore, Pi = Pf , (iv) Example: A marble balanced on top of a i.e., m1u1 + m2u2 = (m1 + m2)v hemispherical bowl. or m1u1 = (m1 + m2)v (u2 = 0) Neutral : m1u1 or v=..(ii) 1 (i) When a particle is slightly displayed from a m1 + m2 position it does not experience any force acting on it and continues to be in equilibrium in Knowing m1, m2 and u1, we can calculate the final the displaced position, it is said be in neutral m1 velocity v. As the mass ratio < 1, v < u1. equilibrium m1 + m2 (ii) Potential is constant  1 d 2U Total K.E. before collision, (iii) =0 dx 2 1 (iv) Example: A marble placed on a horizontal table.5 E1 = mu2 ½ 2 1 1 Q. 2. Derive a relation for an inelastic collision in one dimension. U Total K.E. after collision, Ans. 1 E2 = (m + m2)v2 2 1 2 1  m1u1  E2 = (m1 + m2)   2  m1 + m2  Using (ii) m12u12 ½ or E2 = ½ 2( m1 + m2 ) Figure shows two bodies of masses m1 and m2 moving with velocities u1 and u2 respectively, Difference in K.E. = E1 – E2 along a single axis. They collide involving some loss of kinetic energy. Therefore, the collision is 1 m12u12 = m1u12 – inelastic. Let v1 and v2 be the velocities of the two 2 2( m1 + m2 ) bodies after collision. As the two bodies or one system, is closed and m12u12 + m1m2u12 – m12u12 = isolated, we can write the law of conservation of 2( m1 + m2 ) linear momentum for the two body system as : Total momentum before the collision (pi) m1m2u12 = = Total momentum after the collision (pf) 2( m1 + m2 ) m1u1 + m2u2 = m1v1 + m2v2 ½ In a perfectly inelastic collision, the body of mass m2 The difference being positive it may be concluded happen to be initially at rest (u2 = 0). After the that there is loss in kinetic energy.  1 Objective Type Questions (1 mark each) coefficient of restitution between the surface A Multiple Choice Questions and ball is 0.6, the ball rebounds to a height of (A) 0.6 m (B) 0.36 m Q. 1. A cyclist comes to a skidding stop in 20 m. (C) 3.6m (D) 1m During this process, the force on the cycle due to Ans. Option (B) is correct. the road is 100 N and is directly opposed to the Explanation: The velocity of the ball when it motion. Work done by the road do on the cycle is reaches the ground = v0 = 2gH (A) – 2000 J (B) 2000 J 2g × 1 or, v0 = (C) 1000 J (D) 100 J Ans. Option (A) is correct. ∴ v0 = 2g Explanation: W = Fs cos 180° = 100 × 20 × (–1) = – 2000 J The ball rebounds with a velocity v’ Q. 2. A ball is dropped from a height of 1 m. If the  WORK, ENERGY AND POWER v’ = ev0 = e 2g C Assertion & Reason Type Maximum height travelled by the ball starting Questions 2 vr Assertion (A) is followed by a statement of with a velocity v’ is h = 2g Reason (R). Mark the correct choice as. (A) Both A and R are true and R is the correct e2 × 2 g explanation of A or, h = 2g (B) Both A and R are true but R is NOT the correct explanation of A ∴ h = e2 = 0.36 m (C) A is true but R is false Q. 3. A body of mass 100 g falls from a height of 10 m. (D) A is false and R is true Its increase in kinetic energy is Q. 1. Assertion (A): Mass and energy are not conserved (A) 9800 J (B) 9.8 J separately. They are conserved as a single entity (C) 980 J (D) 100 J called mass-energy. Ans. Option (B) is correct. Reason (R): Mass and energy are inter- convertible. Explanation: Loss of PE = Gain in KE Ans. Option (A) is correct.  100  Explanation: According to Einstein’s theory, mass Loss in PE= mgh =   × 9.8 × 10 = 9.8 J  1000  and energy are related as E=mc2 where, E = Energy So, increase in KE = 9.8 J m =mass Q. 4. 1 joule = 1 ________ c = speed of light (A) Nm (B) Nm2 So, mass and energy are not conserved separately. (C) N m 2 2 (D) N2m They are conserved as a single entity. Hence, assertion and reason both are true and the Ans. Option (A) is correct. reason explains the assertion. Explanation: 1 Joule = 1 Newton × 1 meter = 1 Q. 2. Assertion (A): The kinetic energy of the body of Nm mass 2 kg and momentum of 2 Ns is 1 J. Q. 5. If momentum rises by 20%, kinetic energy rises Reason (R): The relation between kinetic energy by and linear momentum of an object is given by (A) 44% (B) 40% p K=. (C) 88% (D) 20% 2m Ans. Option (A) is correct. Ans. Option (C) is correct. p 2 p2 Explanation: Initially, KE = Explanation: K = 2m 2m p '2 22 Finally, (KE)’ = or, K = 2m 2×2 ∴ K =1J Putting, p’ = 1.20p So, the assertion is true. But the reason is false. (1.20 p )2 Q. 3. Assertion (A): A spring stores potential energy (KE)’ = 2m when compressed as well as when stretched. Reason (R): Potential energy is stored in a body 1.44 p 2 or, (KE)’ = by deformation of its shape. 2m Ans. Option (A) is correct. or, (KE)’ = 1.44 (KE) Explanation: When a spring is compressed or So, KE increases by 44% stretched its shape is deformed. Hence, elastic potential is stored in it. NCERT Corner Q. 1. The sign of work done by a force on a body is (d) Work done by an applied force on a body moving important to understand. State carefully if the on a rough horizontal plane with uniform following quantities are positive or negative : velocity. (a) Work done by a man in lifting a bucket out of (e) Work done by the resistive force of air on a well by means of a rope tied to the bucket. vibrating pendulum is bringing it to rest. (b) Work done by gravitational force in the above Ans. As, W= Fs cos case. (a) = 0, work is positive. (c) Work done by friction on a body sliding down (b) = 180, work is negative. an inclined plane.  CBSE Question Bank Chapterwise & Topicwise, PHYSICS, Class-XI (c) A  s force of friction act opposite to direction of minimum total energy the particle must have motion, work is negative. in each case. Think of simple physical contexts (d) As applied force acts in the direction of for which these potential energy shapes are motion, work is positive. relevant. (e) The resistive force of air acts opposite to V(x) direction of the motion, hence work is negative. V0 Q. 2. A body of mass 2kg initially at rest moves under the action of an applied horizontal force of 7 N E on a table with coefficient of kinetic friction = x 0.1. Compute the O a V(x)=V(a) (a) Work done by the applied force in 10 s, (i) (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, and interpret your results. Ans. Given : m = 2kg, u = 0, F = 7N, m = 0.1, t = 10s F 7 -2 Acceleration = a1= = =3.5ms m 2 (ii) Frictional force = f = µR = µmg          = 0.1  2  9.8          =1.96 N. Retardation- −f −1.96 a2 = = = − 0.98 ms−2 m 2 Net acceleration, a = a1 + a2 = 3.5 – 0.98 (iii)             = 2.52 ms–2 Distance, in 10 seconds 1 s = ut + at 2 2 1 = 0+ ×2.52×(10)2 2 = 126m (a) Work done (applied force) = F.s = 7  126 = 882 J. (b) Work done (frictional force) = −f.s (iv)               = −1.96126 Ans. It is clear that the total energy of the body is given               = −247 J by (c) work done (net force) = Net force  s E = K.E. + P.E. 1 Net force, F = 7 – 1.96 = 5.04 or K.E. = E – P.E. = mv2 2 Work done  = 5.04  126 = 635 J. Hence, K.E. of the body can never be negative. (d) Velocity at the end of 10 second. Thus, P.E. cannot be greater than E. v = u + at = 0 + 2.52  10 = 25.2 m/s. (i) In the region between x = 0 and x = a, P.E. is zero, Initial K.E. = 0 so K.E. is positive. In the region x > a, the P.E. 1 1 (V0) has a value greater than E. So, K.E. will be Final KE= mv 2 = ×2×(25.2)2 = 635J negative in the region. Hence, the particle can’t 2 2 be present in the region x > a. Change in K.E. = 635 −0 = 635J. The minimum total energy that the particle can K.E.’s change is equal to work done by net force. have in this case is zero. (ii) Here, in all the regions, i.e., – ∞ < x < a, a < x < b, Q. 3. Given below are examples of some potential b < x < c, c < x < d and d < x < ∞, the P.E. is energy functions in one dimension. The total greater than the total energy. Hence, the particle energy of the particle is indicated by a cross cannot be present in the region – ∞ < x < ∞. on the ordinate axis. In each case, specify the The minimum total energy that the particle can regions, if any, in which the particle cannot be have in this case is V1. found for the given energy. Also, indicate the  WORK, ENERGY AND POWER (iii) In region x < a and x > b, the P.E. is V0 which gravitational force over every complete orbit of is greater than the total energy of the particle. the comet is zero. Why? So, K.E. will be negative in this region. Thus, the (c) An artificial satellite orbiting the Earth in very particle cannot be present in the region x < a and thin atmosphere loses its energy gradually x > b. In the region between x > a and x < b, the due to dissipation against atmospheric P.E. is negative this clears that there is a positive resistance, however small. Why then does its value of K.E. Hence, the particle can be present in speed increase progressively as it comes closer the region between x > a and x < b. and closer to the Earth? The minimum total energy that the particle can (d) In Figure (i) the man walks 2 m carrying a mass of 15 kg on his hands. In Figure (ii), he have in this case is –V1. walks the same distance pulling the rope behind (iv) In this case, the P.E. of the particle is more than him. The rope goes over a pulley, and a mass b a the total energy (E) in the regions –