Physics. TOPIC : Friction. 1. To avoid slipping while walking on ice, one should take smaller steps because of the

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1 TOPIC : Friction Date : Marks : 0 mks Time : ½ hr. To avoid slipping while walking on ice, one should take smaller steps because of the Friction of ice is large (b Larger normal reaction (c Friction of ice is small (d Smaller normal reaction. If a ladder weighing 50N is placed against a smooth vertical wall having coefficient of friction between it and floor is 0., then what is the maximum force of friction available at the point of contact between the ladder and the floor 75 N (b 50 N (c 5 N (d 5 N. The maximum speed of a car on a road turn of radius 0m; if the coefficient of friction between the tyres and the road is 0.4; will be 9.84 m/s (b 0.84 m/s 4. A block of mass 50 kg slides over a horizontal distance of m. If the coefficient of friction between their surfaces is 0., then work done against friction is 98 J (b 7J (c 56 J 5. A car is moving along a straight horizontal road with a speed v 0. If the coefficient of friction between the tyres and the road is µ, the shortest distance in which the car can be stopped is v0 (b µ g v0 v 0 (c µ g µ g (c 7.84 m/s (d 4 J (d v 0 µ (d 5.84 m/s 6. The coefficient of friction between a body and the surface of an inclined plane at 45 is 0.5. If g = 9.8 m / s, the acceleration of the body downwards in 4.9 m / s is (b 4.9 (c 9. 6 (d Starting from rest, a body slides down a 45 inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is 0. (b 0.5 (c 0.75 (d A block of base 0 cm 0cm and height 0cm is kept on inclined plane. The coefficient of friction between them. The inclination θ of this inclined plane from the horizontal plane is gradually increased from 0. Then At θ =0, the block will start sliding down the plane (b The block will remain at rest on the plane up to certain θ and then it will topple (c At θ = 60, the block will start sliding down the plane and continue to do at higher angles (d At θ = 60, the block will start sliding down the plane and on further increasing θ, it will topple att certain θ Page of 8

2 9. A block is kept on an inclined plane of inclination θ of length l. The velocity of particle at the bottom of inclined is (the coefficient of friction is µ gl ( µ cosθ sinθ (b gl (sinθ µ cos θ (c gl (sinθ + µ cos θ (d gl (cos θ + µ sinθ 0. A body of mass 00 g is sliding from an inclined plane of inclination 0. What is the frictional force experienced if µ =. 7.7 N (b.7 N (c.7 N (d.7 N. A block of mass kg slides down on a rough inclined plane of inclination 60 o starting from its top. If the coefficient of kinetic friction is 0.5 and length of the plane is m, then work done against friction is (Take g = 9.8 m/s 9.8 J (b 4.94 J (c.45j (d.96 J. A body of mass 0 kg is lying on a rough plane inclined at an angle of 0 o to the horizontal and the coefficient of friction is 0.5. the minimum force required to pull the body up the plane is 94 N (b 9.4 N (c 9.4 N (d 0.94 N. 00 Joule of work is done in sliding up a kg block on an inclined plane to a height of 0 metres. Taking value of acceleration due to gravity g to be 0 m/s, work done against friction is 00 J (b 00 J (c 000 J (d Zero o 4. A block rests on a rough inclined plane making an angle of 0 with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If the frictional force on the block is 0 N, the mass of the block (in kg is (take g = 0 m / s.0 (b 4.0 (c.6 5. A block of mass 0. kg is held against a wall by applying a horizontal force of 5 N on the block.. If the coefficient of friction between the block and the wall is 0.5, the magnitude of the frictional force acting on the block is i.5 N (b 0.98 N (c 4.9 N (d.5 (d 0.49 N 6. A body of mass M is kept on a rough horizontal surface (friction coefficient µ. A person is trying to pull the body by applying a horizontal force but the body is not moving. The force by the surface on the body is F, where F = Mg (b F = µ Mgf (c Mg F Mg + µ (d Mg F Mg + µ 7. What is the maximum value of the force F such that the block shown in the arrangement, does not move F 60 m= kg µ = 0 N (b 0 N (c N (d 5 N Page of 8

8. A block of mass kg rests on a rough inclined plane making an angle of 0 with the horizontal. The coefficient of static friction between the block and the plane is 0.7.

3 8. A block of mass kg rests on a rough inclined plane making an angle of 0 with the horizontal. The coefficient of static friction between the block and the plane is 0.7. The frictional force on the block is 9.8 N (b N (c 9.8 N (d N 9. A marble block of mass kg lying on ice when given a velocity of 6 m/s is stopped by friction in 0s. Then the coefficient of friction is 0.0 (b 0.0 (c 0.0 (d A block of mass 00kg is being pulled up by men on an inclined plane at angle of 45 as shown. The coefficient of static friction is 0.5. Each man can only apply a maximum force of 500N. Calculate the number of men required for the block to just start moving up the plane 45 0 (b 5 (c 5 (d. A block at rest slides down a smooth inclined plane which makes an angle 60 with the vertical and it reaches the ground in t seconds. Another block is dropped vertically from the same point and reaches the ground in t seconds. Then the ratio of t : t is : (b : (c : (d : (e :. A coin of mass 0g is placed over a book of length 50 cm. The coin is on the verge of sliding when one nd of the book is lifted 0 cm up. coefficient of static friction between the book and the coin is.0 (b 0.4 (c 0. (d 0.. A body is sliding down an inclined plane having coefficient of friction 0.5. If the normal reaction is twice that of the resultant downward force along the incline, the angle between the inclined plane and the horizontal is 5 o (b 0 o (c 45 o (d 60 o 4. A conveyor belt is moving at a constant speed of m/s. A box is gently dropped on it. The coefficient of friction between them is µ = 0.5. The distance that the box will move relative to belt before coming to rest on it,, taking g = 0ms -, is Zero (b 0.4 m (c m (d 0.6 m 5. A block of mass M = 5 kg is resting on a rough horizontal surface for which the coefficient of friction is 0.. When a force F = 40 N is applied, the acceleration of the block will be ( g = 0 m / s 5.7 m F (b 8.0 m M 0 P a g e

4 (c.7 m (d 0.0 m 6. Consider a car moving on a straightt road with a speed of 00 m/s. The distance at which car can be stopped is [ µ = 0.5] 00 m (b 400 m (c 800 m (d 000 m 7. Two carts of masses 00 kg and 000 kg on horizontal rails are pushed apart. Suppose the coefficient of friction between the carts and the rails are same. If the 00 kg cart travels a distance of 6 m and stops, then the distance travelled by the cart weighing 00 kg is k A B m (b 4 m (c 6 m (d m 8. A motorcyclist of mass m is to negotiate a curve of radius r with a speed v. The minimum value of the coefficient of friction so that this negotiation may take place safely, is v rg (b v gr (c gr v (d g v r 9. Consider a car moving along a straight horizontal road with a speed of 7 km/h. If the coefficient of kinetic friction between the tyres and the road is 0.5, the shortest distance in which the car can be stopped is [ g = 0 ms 0 m (b 40 m (c 7 m (d 0 m ] 0. A body of 0 kg is acted by a force of 9.4 N if the coefficient of kinetic friction 0.0 (b 0.0 (c 0.00 (d 0.5 g = 9.8 m. The acceleration of the block is 0 m / s. What is Page 4 of 8

ANSWERS. (c.. (b 4.. (c. (b. 4.. (b. (d. (c 4. 5. 6. 7. (c 8. (b 9. (b 0. (b 5. (b 6. 7. (c 8. 9. (d 0. (c (b 5. 6. (d 7. (c 8. (b 9. (b 0. (c Solutions. F = µ R = 0. 50 = 75 N. (b v = µ rg = 0.4 0 9.

5 ANSWERS. (c.. (b 4.. (c. (b. 4.. (b. (d. (c (c 8. (b 9. (b 0. (b 5. (b (c (d 0. (c (b (d 7. (c 8. (b 9. (b 0. (c Solutions. F = µ R = = 75 N. (b v = µ rg = = m / s 4. W = µ mgs = = 98 J 5. Retarding force F = ma = µ R = µ mg a = µ g Now from equation of motion v = u 6. a = g(sinθ µ cosθ = 9.8(sin = m 7. (c µ = tan θ n θ = 45 and n = (Given µ = tan 45 = 4 = 8. (b For rotational equilibrium about point P a b as 0 u u u = as s = = a µ µ g o v0 = µ g 0.5 cos 45 = o mg sinθ P θ mg cosθ sin cos.69 i.e., toppling starts at.69 and angle of repose! " 60 it mean the block will remain at rest on the plane up to certain angle θ and then it will topple. 9. (b Acceleration = g(sin θ µ cosθ and s = l 0. (b F = µ R µ mg cosθ k k = v = as = gl(sinθ µ cos θ k 5 P a g e

6 F = cos 0 =. 7 k. (c W = µ mg cos θ S = =.45 J. (b F = mg (sinθ + µ cos θ = 0 9.8(sin cos 0 = 9.4 N.. Work done against gravity = mgh Work done against friction = (Total work done work done against gravity = = 00 J 4. Angle of repose = tan α ( µ = tan (0.8 = 8.6 Angleof inclined plane is given θ = 0. It means block is at rest therefore, 5. (b Limiting friction Since downward force is less than friction will work on it. F s = downward force = Weight = = 0.98 N N = 0 0 = 00 J 0 Static friction = component of weight in downward direction = mg sin θ = 0 N m = = kg 9 sin 0 R µr mg 5N F = µ sr = 0.5 (5 =. l 5 N limiting friction therefore block is at rest so the static force of 6. R F f 60 F cos 60 f = µr F sin 60 F cos 60 = µ ( W + F sin 60 W = 0 Substituting µ = & W = 0 we get F = 0 N 7. (c Maximum force by surface when friction works F = f + R = ( µ R + R = R µ + Minimum force = R when there is no Hence ranging from R to R µ + We get, Mg F Mg µ + 8. Limiting friction F l = µ mg cosθ friction F l = cos 0 = N (approximately But when the block is lying on the inclined plane then component of weight down the plane = 9.8 sin 0 = 9.8 N It means the body is stationary, so static friction will work on it Static friction = Applied force = 9.8 N Page 6 of 8 = mg sin θ

7 u 6 9. (d v = u at u µ gt = 0 µ = = gt (c R NF = gsin gcos45 00 Here, mass of the block, m =00kg Coefficient of static friction, µ $ 0.5 angle of inline plane, θ + µsr 00 % 0'sin45 * cos % 0' * + % - %%. 5. % %.. (b H 60 L Let L be the length and H be height of the inclined plane respectively Acceleration of the block slide down the smooth incline plane is a = g cos 60 / 060 Acceleration of another block dropped vertically down from the same inclined plane is From the same inclined plane is a = g From figure, cos 60 = 4 cos60 5 /060 Divide (i by (ii, we get :$; <=>; :$ 8 ;? AB 0C AB 0C 7 P a g e

8 . Tangent of angle of repose = Coefficient of static friction $ tan $ F F 0... (c Resultant downward force along the incline = mg (sin θ µ cos θ Normal reaction = mg cos θ Given : mg cos θ = mg (sinθ µ cos θ By solving o θ = (b 5 5. G B * 0 * % 5 H I.J..KLM or distance =0.4 m R F sin 0 F Fk 0 F cos 0 6. (d Kinetic friction = µ = 0.( mg F sin 0 R k = = 0.( 50 0 = 6 N Acceleration of the block = F cos 0 Kinetic friction Mass 40 = 5 6 = 5.7 m / s u (00 s = = = 000 m µ g s 7. (c For given condition s m s 4 4 s = s = 6 = 6 m 9 9 u (0 9. (b s = = = 40 m µ g NOP 7 % 0/ QR T 0. (c Net force on the body = Applied force Friction ma = F µ mg k µ k mg m = m F ma = = = 0. mg = 00 Page 8 of 8

Mechanics II. Which of the following relations among the forces W, k, N, and F must be true?

Mechanics II. Which of the following relations among the forces W, k, N, and F must be true? Mechanics II 1. By applying a force F on a block, a person pulls a block along a rough surface at constant velocity v (see Figure below; directions, but not necessarily magnitudes, are indicated). Which