PHYSICS LAWS OF MOTION CLASS: XI

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1 In this chapter we will study the cause of motion of bodies, the factors affecting motion of body such as force, friction, mass etc. The branch of physics which deals with the study of motion of object with taking into account the factors causing its motion is called dynamics. INERTIA We see that a body at rest tends to remain at rest and do not starts moving by itself. For example a chair, a car or any other object at rest do not move itself until we apply some effort to move it. Reverse is also true. if a body is in motion then it tries to keep its motion continuous, provided there is no frictional force or retarding force. Def n : This property of body due to which it is not able to change its state of rest or of uniform motion along a straight line. Inertia is divided in three categories: (a) Inertia of rest: It is the property of a body due to which it is not able to change its state of rest on its own. For example: 1. When a bus starts suddenly then passenger falls backwards, this is due to the inertia of rest of the upper potion of bodies of passengers.. When a horse starts suddenly then rider falls backwards, this is again due to the inertia of rest of upper portion of body of horse rider. (b) Inertia of motion: It is the property of a body due to which it is not able to change its state of uniform motion along straight line on its own. For example: 1. When brakes are applied suddenly to a moving bus then passengers falls forward, this is due to the inertia of motion of upper portion of bodies of passengers.. When a moving horse stops suddenly then horse rider falls forward, this is again due to inertia of motion of the upper portion of body of horse rider. (c) Inertia of direction: It is the property of a body due to which it is not able to change its direction of uniform motion along straight line. For example: 1. When a bus takes sharp turn, then passengers are thrown outwards, this is due to the inertia of direction of upper portion of bodies of passengers.. When a stone tied to one end of a string is whirled and the string breaks suddenly, the stone flies off along the tangent to the circle. This is because the pull in the string was forcing the stone to move in a circle. As soon as the string breaks. The stone tends to move along straight line, and flies off tangentially. FORCE Force is the external effort applied to a body which changes or tries to changes its state of rest or state of uniform motion along straight line or direction of motion. NEWTON S FIRST LAW It states that. a body will tends to remain in a state of rest or of uniform motion along straight line, until it is compelled by some external force to change that state. A.K Maggo ; M.Tech [IITD]; (M)

2 Newton s First law defines the property of inertia of body. Hence all practical examples given for inertia are applicable to Newton s First law. LINEAR MOMENTUM Linear momentum of body is the quantity of motion contained in a body. [Denoted by p] Quantity of motion of body depends upon two factors (a) It is directly proportional to mass of body p m (b) It is directly proportional to velocity of body Combining above two factors we have p v p mv p mv Here constant of proportionality is 1. Def n : Thus Linear momentum of a body is defined as the product of its mass and velocity. Linear momentum is vector quantity and its direction is always along the direction of velocity of body. SI units for linear momentum are kgms -1. Its dimensions are MLT -1 NEWTON S SECOND LAW It states that the rate of change of linear momentum of a body is directly proportional to force applied on body. dp i.e. F dp F, Here constant of proportionality is 1. Explanation of Newton s Second law Consider a body of mass m, moving with a uniform acceleration a along straight line. According a Newton s second law, dp F Putting p = mv, we get, d( mv) F d( v) F m d( v) F ma[ a] Hence F = ma. Thus Force acting on a body is defined as the product of its mass and its acceleration, in accordance with Newton s second law. A.K Maggo ; M.Tech [IITD]; (M)

3 SI UNIT OF FORCE SI unit of force is Newton [N] Consider a body of mass, m = 1kg, accelerating with acceleration of 1 m/s, then force acting on it is given by F = 1kg 1m/s or F = 1N Def n : One Newton is that force which provides an acceleration of 1 m/s in a body of mass 1 kg. CGS UNIT CGS unit of force is dyne. Consider a body of mass 1 g, moving with acceleration of 1 cm/s along straight line, then force experienced by it is: F = 1g 1cm/s or F = 1 dyne. Def n : One dyne is that force which when applied on a body of mass 1 g provides it an acceleration of 1 cm/s. Gravitational units: MKS Gravitational unit of force is Kgf or kgwt, 1 Kgf = 9.8 N CGS gravitational unit of force is gwt or gf, 1 gwt = 980 dynes. SOME RESULTS FROM NEWTON S SECOND LAW 1.Inertial mass: From Newton s second law we have F F = ma, thus m, Hence Inertial mass of a body is defined as the ratio of force applied on a body a to its acceleration produced due to that force..no Force is required to move a body uniformly: Consider a body moving with uniform velocity, then its acceleration will be zero, as velocity is constant, thus by Newton s Second law F = m 0, F = 0. 3.On the basis of Newton s Second law we can measure force applied on a body as the product of its mass and rate of change of velocity [acceleration.] IMPULSE During an impact such as hitting a ball with a bat, falling of a body on a floor, or collision of two bodies, a large amount of force acts on body for a very small interval of time. Such forces which acts on bodies for very small interval of time are called impulsive forces. Def n : Impulse of a force on a body is defined as the product of the force of impulse experienced by body and the time for which impulsive force acts on body, Impulse = Force time. Impulse is denoted by I. Thus I = F t (1) EXPRESSION FOR IMPULSE According to Newton s second law we have dp F A.K Maggo ; M.Tech [IITD]; (M)

4 F dp, i Integrating both sides. t 0 F F t 0 p p1 dp p p1 dp F[ t] p p From equation 1, we have, I = F t = p p 1. Thus impulse is also defined as the net change in momentum of a body, during an impact. SI unit: SI unit of impulse is Ns Dimensions: MLT -1. Note: The impulsive force experienced by body is not constant but it changes with time as shown in following graph. The impulse of a force is also defined as the area enclosed by the graph with time axis. Thus, impulse = area of curve. F 1 t APPLICATIONS OF IMPULSE 1. A person experiences more injury when it falls on a cemented floor as compared to sandy ground. When person falls on a sandy ground, time of collision is more as compared to a cemented floor, by dp second law F, as is large F will be small and person experiences less injury.. A cricket player lowers its hands while catching the ball to avoid injury: By lowering its hands, the cricketers increases the time of impact due to which force of impact F dp decreases in accordance with Newton s second law F 3. Bogies of a train are provided with spring system: When a jerk or a shock occurs in the way of train, springs increases the time of shock, to reach the passengers. Since increases, hence force exerted decreases, and passengers experiences less injury. A.K Maggo ; M.Tech [IITD]; (M)

5 PRINCIPLE OF CONSERVATION OF LINEAR MOMENTUM It states that for an isolated system, consisting of any number of particle, vector sum of linear momentum of all the particles remains conserved if no external force is applied. Proof: Consider a system of n particles in which different bodies have linear momentum p, p p Total linear momentum of system By Newton s second law net force acting on system is given by dp F no external force acts so F = 0 dp 0 dp 0 P cos ntant From (1) p p p pn cons tant P 1, p1 p p3 p n (1) 1 3 Hence proved. Note: According to principle of conservation of linear momentum, net momentum of system is conserved in the absence of any external force thus we always have [Total initial momentum] = [ Total final momentum] i.e. m 1 u 1 + m u = m 1 v 1 + m v. [ For a system of two particles.] RECOIL OF A GUN When we fire a bullet from a gun, it is observed that gun recoils back. If we know the velocity of bullet fired then we can easily calculate the recoil velocity of gun. Consider a gun of mass m, consisting of a bullet of mass m 1. If bullet is fired with velocity v 1 then say gun recoils back with velocity v. 3 p n V V 1 Initially both the bullet and gun are at rest before fire. initial total linear momentum of system m 1 u 1 + m u = = (1) A.K Maggo ; M.Tech [IITD]; (M)

6 After firing the bullet, final linear momentum of system. By conservation of linear momentum : m 1 v 1 + m v () 0 = m 1 v 1 + m v m1 v v1 m Here negative sign shows that gun recoils in opposite direction to the bullet. NEWTON S THIRD LAW: It states that to every action there is always equal and opposite reaction. Suppose bodies A and B are in contact with each other. A exerts a force F AB on B, and B exerts a force F BA on A as reaction. According to Third law action and reaction are equal and opposite, thus F AB = - F BA NORMAL FORCES OR CONTACT FORCES Whenever two bodies are in contact with each other they exert contact forces on each other, by Newton s third law these forces are equal and opposite. Normal reaction: If a body is placed on a surface, then due to its weight it exerts some force on surface. Then surface in accordance with Newton s third law proved reaction in normal direction, this normal force is always perpendicular to surface. e.g. a body is placed on a horizontal surface, exerts a force W = mg, equal to its weight on the surface. Then surface provides a normal reaction R perpendicular to it and opposite to W as shown. Now R = W Hence R = mg. R mg Note: If m is the mass of a body then its weight, is given as the product of m and g, i.e. mg. Weight of body always acts vertically downward. A.K Maggo ; M.Tech [IITD]; (M)

7 DERIVATION OF NEWTON S FIRST LAW FROM NEWTON S SECOND LAW By Newton s Second law we have, dv F m, if no force is applied on a body then, F = 0, dv Thus 0 m 0 dv v cons tant This shows that if no force is applied on a body then, its velocity remains constant, i.e. do not change. It implies that if body is at rest then it will be at rest and if it is moving with constant velocity along straight line, the it will continue its motion until F = 0. This proves 1 st law. DERIVATION OF THIRD LAW ON THE BASIS OF SECOND LAW. Consider two bodies in contact with each other F AB A B F BA F AB = Force exerted by body A on B. F BA = Force exerted by body B on A. Δp 1 = change in linear momentum of body A Δp = change in linear momentum of body B By Newton s Second law p1 F AB (1) t p And F BA () t Since system is isolated no external force acts on system and hence net momentum is conserved, and net change in momentum of system is zero. p p 0 Putting values from (1), and () F AB t 1 F BA t 0 A.K Maggo ; M.Tech [IITD]; (M)

8 F AB F BA F 0 AB F BA Hence proved. TENSION Suppose a body connected with a light string or a rope as shown, now if string is pulled with a force of 10 N [say], then body experiences same for of 10 N. It means that force exerted on rope is transmitted to the body. Such forces which are transmitted by rope or a string to other bodies connected with it are called forces of tension. Remember we can only pull a string but we cannot compress it, hence force of tension is always pull type, i.e. away from the body in contact, as shown below. T Note: If string is light and flexible, then tension is same throughout its length, i.e. if T is the tension at one end, the on other end it will be T. Q. What is Tension, in the string and acceleration of the system of masses m 1 and m connected to each other, as shown, assume string and pulley to be massless. APPARENT WEIGHT OF A MAN IN A LIFT When you stand in a lift, and lift accelerates in upward direction, you feel slightly heavier. When lift accelerates in downward direction, you feel slightly lighter than your original weight. This weight that a body experiences during the upward or downward motion of a lift is called apparent weight. Case-1: When lift is accelerating downward: In this case acceleration of lift is downward. Force equation body is mg R ma R mg a Here R is the normal reaction provided by the floor of lift, this is equivalent to apparent weight of body. Thus, R mg ma R W ma, where, W is the actual weight of body. A.K Maggo ; M.Tech [IITD]; (M)

9 Thus apparent weight of body is less than its actual weight during downward acceleration of lift. Case-: When lift is accelerating upwards: In this situation force equation of body R mg ma R mg ma, Thus apparent weight of body is greater than its actual weight. R m( g a) Case-3: When lift is at rest: R In this case force equation of body, R mg Thus apparent weight is equal to real weight when lift is at rest. Case-4: When lift is moving uniformly, i.e. with uniform velocity. In this case also lift has no acceleration, i.e. a = 0, force equation of body. R mg Hence apparent weight is equal to real weight. mg Case-5: When lift is falling freely: In this case downward acceleration of lift is g, i.e. a = g. Thus force equation of lift is mg R mg R 0 Thus apparent weight is zero, i.e. body experiences weightlessness. FRICTION When a body slides or tries to slide over a rough surface there exists some opposing force which opposes the motion of body. Thus friction is the opposing force which comes into play whenever a body moves or tries to move over the surface of another body. Friction is of two types, sliding friction and rolling friction. Sliding friction: It is the opposing force which comes into play whenever a body slides or tries to slide on the surface of another body. Rolling friction: It is the opposing force which comes into play whenever a body rolls or tries to roll on the surface of another body. A.K Maggo ; M.Tech [IITD]; (M)

10 CAUSE OF FRICTION Classically it was assumed that the roughness of surfaces is the only cause of friction. But according to modern point of view friction is observed as the opposing force arising from the molecular forces acting between surfaces. This is assumed as there is friction between the smooth surfaces also. TYPES OF SLIDING FRICTION Consider a solid object placed on a rough horizontal table. Apply some force in horizontal direction, and increase slowly its value. It is observed that initially block does not move. This is due to the frictional force acting between horizontal surface and block. Thus applied force is cancelled by the frictional force and body remains at rest, such frictional force which acts on a body at rest is called static friction. When force is increased to certain limit body starts sliding down on the surface. That limit of sliding friction above which motion of body just starts is called limiting friction. When body is in motion, frictional force still acts on it in opposite direction. That frictional force which acts on a body in motion is called sliding friction. Static friction: It is the opposing force which comes into play when a body tries to move on the surface of anther body. Note: Static friction is self adjusting in nature, i.e. it balances applied force. Sliding friction is always equal to the applied force on the body. Limiting friction: It is the maximum opposing force which comes into play when a body tries to move on the surface of another body but actual motion is just begin to start. Limiting friction is the maximum of static friction. Kinetic friction: It is the opposing force which comes into play when a body actually slides on the surface of another body. GRAPH BETWEEN APPLIED FORCE AND FRICTIONAL FORCE Consider an object placed on a horizontal surface. Apply some horizontal force F on it, and increase its value slowly. First static friction will act, then limiting friction and finally kinetic friction as discussed previously. Following is the graph obtained. Here OA, represent the static friction curve. At point A friction is maximum and is representing limiting friction. At point B body starts sliding and is representing kinetic friction. A.K Maggo ; M.Tech [IITD]; (M)

11 F F f f LAWS OF LIMITING FRICTION: 1. Force of limiting friction acting on a body placed on a rough surface is directly proportional to the normal reaction provided by the surface. i.e. f R f R, Where µ, is constant of proportionality and is called coefficient of friction. Value of µ depends upon the relative roughness of surfaces. Greater be the roughness of surfaces greater will be µ. Generally it is less than one. Def n of µ: We have f R, Thus coefficient of friction between any two surfaces is defined as the ratio of the force of friction experienced by a body placed on it to the normal reaction provided by the surface to that body. It is a unitless quantity.. Force of limiting friction between any two surfaces depends upon the roughness of surfaces, greater be the relative roughness greater will be force of limiting friction. 3. Force of limiting friction between any two surfaces is independent of the apparent area of contact of bodies. 4. Force of limiting friction can be reduced by using lubrication, or using ball bearings. ANGLE OF FRICTION It is the defined as the angle that the resultant(r ) of force of limiting friction(f) and normal reaction(r) forms with the direction of normal reaction. It is denoted by θ. From right triangle OAB we have, R R f f tan (1) R A.K Maggo ; M.Tech [IITD]; (M)

12 Using law of limiting friction, f = µr () From (1) and () Tanθ = µ Hence tangent of angle of friction is equal to the coefficient of limiting friction between the surfaces. CIRCULAR DYNAMICS Uniform circular motion: In this type of motion a body moves with a uniform speed along a circular track. The acceleration experienced by body in uniform circular motion is called centripetal acceleration. It is denoted by a c and is given as: v a c r a c r Here v = speed of body, r = radius of circular path, ω = angular velocity of body. The force required to produce centripetal acceleration is called centripetal force. CENTRIPETAL FORCE It is the force required to move a body uniformly along a circular path, its direction is always toward the centre of circle. This force is also called radial force. It is given as Fc ma c mv F c r F c mr CENTRIFUGAL FORCE Due to the inertia of direction, a body moving uniformly along a circular path, experiences a force that actually tends to bring the body to move along straight line. Thus centrifugal force is the force that acts on a body due to its inertia of direction; this is also called a pseudo force. Centripetal force and centrifugal forces acts on different body (Like action and reaction forces). Centripetal and centrifugal forces being the forces of action and reaction act always on different bodies. Hence they are always equal and opposite. mv F g r A.K Maggo ; M.Tech [IITD]; (M)

13 F g mr BANKING OF ROADS During a turn a vehicle requires some centripetal force for turning. To avoid any skidding or accident, roads are banked at sharp turns. Def n : The phenomenon of raising the outer edge of a road over the inner edge is called banking of roads. Suppose a road is banked by an angle of θ. A vehicle of mass taking turn over this road, experiences two forces: (1) Weight, = mg, in vertically downward direction and () Normal reaction or contact force provided by the road in perpendicular direction as shown, = R. Rcosθ R Rsinθ mg Resolving R into rectangular components we have: Rcosθ (opposite to mg) Rsinθ (towards the centre of circular turn) Here Rcosθ balances the weight of vehicle, and Rsinθ is used to provide necessary centripetal force for turning. Thus Dividing () by (1) we have, mv Rsin (1) r Rcos mg () v tan rg If µ is the coefficient of friction between the tires of the vehicle and the road then, maximum velocity with which vehicle cross the turn is given as v rg (tan ) 1 tan A.K Maggo ; M.Tech [IITD]; (M)

14 BENDING OF A CYCLIST While taking a sharp turn at high speed, a cyclist has to bend toward the centre of circular turn; this is called bending of cyclist. y bending itself, cyclist provides centripetal force for turning. Rsinθ R Rsinθ mg Consider diagram, here R is the normal reaction provided by the surface of road to cycle. Suppose θ is the angle of bending. Resolving R into rectangular components we have: Rcosθ (opposite to mg) Rsinθ (toward the centre of circular turn) Rcosθ component balances the weight of cyclist, and R sinθ provides necessary centripetal force for turning. Dividing () by (1) we have, mv Rsin (1) r Rcos mg () v tan rg In this way, by knowing, r, and v we can find the angle of bending. ROUNDING OF A LEVEL ROAD If a road is not banked at the place of circular turn, then necessary centripetal force required for turning is provided by frictional force acting between the tires of vehicle and ground. The required centripetal force should be less than the available friction; otherwise vehicle will skid, and may lead to an accident. Thus F c f F c R mv r v rg mg This is the maximum safe velocity with which vehicle should take turn. A.K Maggo ; M.Tech [IITD]; (M)

15 F f A.K Maggo ; M.Tech [IITD]; (M)

16 NUMERICALS 1. Fuel is consumed at the rate of 100 kg/s in rocket. The exhaust gases re ejected at a speed of m/s w.r.t the rocket. What is the thrust experienced by the rocket? Also calculate the velocity of the rocket at the instant, when its mass is reduced to 1/10 th of its initial mass. Take initial velocity of the rocket to be zero. [ N, m/s]. A heavy box of mass 0 kg is placed on horizontal surface. If coefficient of kinetic friction between the box and the horizontal surface is 0.5, calculate the force of kinetic friction. Also, calculate acceleration produced under a force of 98 N applied horizontally? [49 N,.45 m/s ] 3. Find the distance traveled by a body before coming to rest, if it is moving with a speed of 10 m/s and the coefficient of friction between the ground and body is 0.4 [1.75 m] 4. A block of mass kg is placed on the floor. The coefficient of static friction is 0.4. A force of.5 N is applied on the block as shown the fig. Calculate the force of friction between the block and the floor. [.5 N] 5. A bullet of mass 0.01 kg is fired horizontally into a 4 kg wooden block at rest on a horizontal surface. The coefficient of kinetic friction between the block and the surface is 0.5. The bullet remains embedded in the block and the combination moves 0 m before coming to rest. With what speed did the bullet strike the block? [ m/s] 6. A mass of 00 kg is placed on a rough inclined plane of angle 30 deg. If coefficient of limiting friction is 0.577, find the greatest and the least forces in Newton acting parallel to the plane to keep the mass in equilibrium. [0, 1960 N] 7. A block of mass 10 kg is sliding in a surface inclined at an angle of 30 0 with the horizontal. Calculate the acceleration of the block. The coefficient of kinetic friction between the block and the surface is 0.5. [0.657 m/s ] 8. When an automobile moving with a speed of 36 km/hr reaches an upward inclined road of angle 30 deg. Its engine is switched off. If the coefficient of friction is 0.1, how much distance will the automobile cover before coming to rest? Take g = 10 m/s. [8.53 m] 9. A block slides down an incline of With the horizontal. Starting from rest, it covers 8 m in the first two sec. Find the coefficient of kinetic friction between the two. [0.11] 10. A block of mass kg rests on a plane inclined at 30 0 with the horizontal. The coefficient of friction between the block and the surface is 0.7. What will be the frictional force action on the block? [11.9 N] IMP. NCERT QUESTIONS A stone of mass 0.5 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with speed of 40 rev/min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 00 N? [34.6 ms -1 ] 1. In above problem if the speed of the stone is increased beyond the maximum A.K Maggo ; M.Tech [IITD]; (M)

17 permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks. (i) The stone jerks radially outwards. (ii) The stone flies off tangentially from the instant the string breaks. (iii) The stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle. 13. A helicopter of mass 1000 kg rises with a vertical acceleration of 15 ms -. The crew and the passenger weigh 300 kg. Give the magnitude and direction of (a) force on the floor by the crew and passengers (b) action of the rotor of the helicopter on the surrounding air (c) force on the helicopter due to the surrounding air 14. An air craft executes a horizontal loop at a speed of 70 km/hr with its wings banked at What is the radius of the loop? [15. km] 15. A train rounds an unbanked circular bend of radius 30 m at a speed of 54 km/h. The mass of the train is 10 6 kg. What provides the centripetal force required for this purpose? The engine or the rails? What if the angle of banking required to prevent wearing out of the rails? [36 0 ] 16. A monkey of mass 40 kg climbs on a rope which can stand a maximum tension of 600 N. In which of the following cases will the rope break. The monkey: (a) Climbs up with an acceleration of 6 m/s. (b) Climbs down with an acceleration of 4 m/s. (c) Climbs up with a uniform speed of 5 m/s. (d) Falls down the rope nearly freely under gravity. 17. A block of mass 15 kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is The trolley accelerates from rest with 0.5 ms - for 0 s and then moves with uniform velocity. Discuss the motion of the block as viewed by (a) a stationary observer on ground (b) an observer moving with trolley. 18. A disc revolves with a speed of 33.5 rev/min. and has a radius of 15 cm. two coins are placed at 4 cm and 14 cm away from the centre of the record? If the coefficient of friction between the coins and the record is Which of the coin will revolve with the record? [coin at 4 cm will move with disc] CONCEPTUALS 19. Name one factor on which the inertia of a body depends. 0. Why does a swimmer push the water backward? 1. Shall man experience a backward jerk when firing a bullet from a gun?. Rocket works on the principle of conservation of Fill in the blanks. 3. A body of mass m is moving with constant velocity v on a horizontal surface. What is the force on the table? 4. If you jerk a piece of paper from under a book quick enough, the book will not move. Why? 5. Why bodies of small mass require small initial effort to put them in motion? 6. Why passengers are thrown outwards when a car in which they are traveling suddenly takes a circular turn? 7. Why is it difficult to walk on a slippery ground? A.K Maggo ; M.Tech [IITD]; (M)

18 8. A body is moving with uniform velocity. Can it be said to be in equilibrium? 9. A stone, when thrown on a glass window, smashes the window pan to pieces. But a bullet fired from a gun p[asses through making a hole Why? 30. If you place a brick on a smooth floor, you can move it easily by pushing it with your foot. Yet if you kick the brick, you will hurt your foot. Why? 31. A body of mass 0 g is moving with a constant velocity of 3 m/s on a horizontal frictionless surface in a vacuum. What is th4 relation between the directions of motion of the two fragments? 3. What is the effect on the acceleration of particle if the net force on the particle is doubled? 33. Stat principal of conservation of linear momentum. 34. From which Newton s laws of motion, the definition of force comes? 35. A bus weighing 900 kg is at rest on the bus stand. What is the linear momentum of the bus? 36. Why do we beat dusty blanket with a stick to remove dust particles? 37. According to Newton s third law, every force is accompanied by an equal in magnitude and opposite reaction. Then how can a movement ever take place? 38. Why do you fall forward when a moving train decelerates to stop and fall backward when a train accelerates from rest? What would happen if the train rounded a curve at constant speed? 39. A masless rope is hung over a frictionless pulley. A monkey holds on to one end of the rope and mirror, having the same light as the monkey, is attached to the other end of the rope at monkey s level. Can the monkey get away from his image seen in the mirror (a) by climbing up the rope (b) by climbing down the rope (c) by releasing the rope? 40. Why a parachute descends slowly? 41. Why is it easier to roll a gas cylinder than to pull it along the road? 4. Why sand is sometimes thrown on the railways tracks in rainy days? 43. Friction is a self adjusting force. Is this statement correct? 44. a body moving over the source of another body suddenly comes to rest. What happens to friction between the two surfaces? 45. Why force of friction increases when the two surfaces in contact are made extremely smooth? 46. What do you mean by dry friction and liquid or fluid friction? 47. A wooden box is lying on an inclined plane. What is the coefficient of friction if the box starts sliding when the angle of inclination is 30 0? A.K Maggo ; M.Tech [IITD]; (M)