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Laws Of Motion Test - 16

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Laws Of Motion Test - 16
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  • Question 1
    1 / -0

    A block of mass 2 kg is placed on the floor. The coefficient of static friction is 0.4. If a force of 2.8 N is applied on the block parallel to floor, the force of friction between the block and floor is (Take g = 10 m/s2)   

    Solution

    Weight of the block is 20N thus maximum friction that can act is 0.4 x 20 = 8N
    But as the external force acting, F = 2.8N < 8N,
    We get f = F = 2.8N

  • Question 2
    1 / -0

    A block has been placed on an inclined plane. The slope angle θ of the plane is such that the block slides down the plane at a constant speed. The coefficient of kinetic friction is equal to  

    Solution


    At a constant speed means, acceleration is 0. Therefore, using formula
    a = g( sinθ - ucosθ)
    0 = g(sinθ - u cosθ)
    ucosθ = sinθ
    u = sinθ/cosθ
    u= tanθ
    Hence D is correct.

  • Question 3
    1 / -0

    A body of mass 60 kg is dragged with just enough force to start moving on a rough surface with coefficients of static and kinetic frictions 0.5 and 0.4 respectively. On applying the same force, what is the acceleration?  

    Solution

  • Question 4
    1 / -0

    `A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab. The static coefficient of friction between the block and the slab is 0.60 while the kinetic coefficient is 0.40. The 10 kg block is acted upon by a horizontal force of 100 N. If g = 9.8 m/s2 the resulting acceleration of the slab will be  

    Solution

    Block will slip on the slab as F > flim Hence kinetic friction acts between block and slab,

  • Question 5
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    If the coefficient of friction of a surface is μ = √3, then the angle of inclination of a plane to make a block on it to just start sliding is

    Solution

  • Question 6
    1 / -0

    If A and B moves with acceleration ‘a’ as shown in diagram, calculate acceleration of C with respect to B 

    Solution


  • Question 7
    1 / -0

    A block takes twice as much time to slide down a 45o rough inclined plane as it takes to slide down a similar smooth plane. The coefficient of friction is   

    Solution

  • Question 8
    1 / -0

    The acceleration that is produced by a 15N force in a mass of 8 kg will be equal to

    Solution

    As we know that F = ma
    20 = 8a
    => 20/8 = a
    a = 1.87 ms-2

  • Question 9
    1 / -0

    A block of mass 6 kg is being pulled by force 24 N as shown. If coefficient of friction between block and the surface is 0.6, frictional force acting on the block is   

    Solution

  • Question 10
    1 / -0

    In figure, coefficient of friction between m1 and m2 is m and that between m1 and the wall is zero. A force F is pressing the system against the wall. Minimum value of force required to hold the system in equilibrium is 

    Solution

    There is no force which can balance the weight of m1.  

  • Question 11
    1 / -0

    A body moves along a circular path of radius 5 m. The coefficient of friction between the surface of the path and the body is 0.5. The angular velocity in rad/s with which the body should move so that it does not leave the path is (g = 10 m/s2)  

    Solution

  • Question 12
    1 / -0

    In figure, coefficient of friction between m2 and the horizontal surface is μ, minimum value of m so that the system has zero acceleration, is  

    Solution

    For zero acceleration,

     Putting force = Frictional force   m1g = μ(m2 + m)g 

  • Question 13
    1 / -0

    A uniform rope of length l lies on a table. If the coefficient of friction is μ, then the maximum length l1 of the part of this rope which can overhang from the edge of the table without sliding down is   

    Solution

  • Question 14
    1 / -0

    A man of mass m stands on a platform of equal mass m and pulls himself by two ropes passing over pulleys as shown in figure. If he pulls each rope with a force equal to half his weight, his upward acceleration would be 

    Solution

  • Question 15
    1 / -0

    The car A is used to pull a load B with the pulley arrangement shown. If A has a forward velocity v determine an expression for the upward velocity vB of the load in terms of vA and θ.θ is angle between  string and horizontal   

    Solution

    Let θ angle between string and horizontal, T is tension in string

  • Question 16
    1 / -0

    A man slides down on a telegraphic pole with an acceleration equal to one-fourth of acceleration due to gravity. The frictional force between man and poles is equal to (in terms of man’s weight w)

    Solution

  • Question 17
    1 / -0

    A cart of mass M has a block of mass m attached to its as shown in the figure. The coefficient of friction between the block and cart is μ. The minimum acceleration of the cart so that the block m does  not fall is

    Solution

    Let the acceleration of the cart be a, then normal reaction on cart is N = ma and frictional force between
    block and cart is f = μN = μma.

    For equilibrium, μma = mg ⇒ a = g/μ

  • Question 18
    1 / -0

    A block is placed on a rough surface and an external horizontal force F is applied on it. The force of friction ‘f’ exerted by the floor on the block is measured for different values of F. The graph between F and f is  

    Solution

    Theoretical  

  • Question 19
    1 / -0

    If the coefficient of friction between an insect and bowl surface is μ and the radius of the bowl is r, the maximum height to which the insect can crawl in the bowl is  

    Solution

  • Question 20
    1 / -0

    A block rests on a rough inclined plane making an angle of 30o 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 10 N, the mass of the block (in kg) is (Take g = 10 m/s2)  

    Solution

    mg sin θ = 10
    m x 10 x (1/2) = 10
    m = 2 kg

  • Question 21
    1 / -0

    Pulleys and strings are massless. The horizontal surface is smooth. What is the acceleration of the block   

     

    Solution

    T = ma,  F = 2T  

  • Question 22
    1 / -0

    A ball of mass 0.5 kg is attached to the end of a string having length 0.5 m. The hall is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of angular velocity of ball in (rad/s)  

    Solution

    T sin θ = mrω2;  r = l sin θ

  • Question 23
    1 / -0

    In the pulley-block arrangement shown in figure, find the relation between acceleration of block A and B  

    Solution

    x1 + x3 = l

    Differentiating with respect to time,

      we get,  v1 + v3 = 0

      Again differentiating with respect to time, 

    Differentiating with respect to time,  v1 + v2 – 2v3 = 0  

    Again differentiating with respect to time, a1 + a2 – 2a3 = 0  

  • Question 24
    1 / -0

    Frictional force between two bodies  

    Solution

    Friction does not always oppose motion. For eg when a car is accelerating forward , the tyres push against the road in a backward direction. Friction pushes itself in forward and the car also pushes forward. Thus friction and motion are in the same direction.
    Similarly when a runner accelerates forward the shoe pushes backward, friction pushes forward and runner pushes forward. Thus both are in same direction
    Hence force of friction does not always oppose motion or the tendency of motion.

  • Question 25
    1 / -0

    The limiting friction between two surface does not depend  

    Solution

    Conceptual  

  • Question 26
    1 / -0

    A body is sliding down a rough inclined plane of inclination 60o. If the angle of repose is 30o, the acceleration of the body (g = 10 m/s2) in m/s2 is  

    Solution

    Angle of repose is the maximum angle of inclination when the downward force on the block is exactly balance by the frictional force. 

    i.e. F = μR = μmgcosθ = mgsinθ, where, θ is the angle of repose. 

    Solving this we will get:

    μ = tan300 = 1/(3)1/2.

    Now when it will start moving then the acceleration of the box will be:

    a = g(sinα−μcosα)

    where α=60o

    ∴  a=10(0.866 − 1/2(3)½]=5.77 m/s2

  • Question 27
    1 / -0

    A cyclist of mass 30 kg exerts a force of 250 N to move his cycle. acceleration is 4 ms−2. force of friction between road and tyres will be

    Solution

    F = ma

    so a = 250/30 

    a = 8.34

    now the result is 4mt/secsqure

    so reduction is 4.34

    friction force will b 4.34 ×30 

     F = 130.2 N

  • Question 28
    1 / -0

    A conveyor belt is moving at a constant speed of 2 m/s. A box is gently dropped on it. The coefficient of friction between them is m  = 0.5. The distance that the box will move relative to belt before coming to rest on it, taking g = 10 m/s2, is  

    Solution

  • Question 29
    1 / -0

    A block of weight 100 N is pushed by a force F on a horizontal rough plane moves with an acceleration 1 m/s2, when force is doubled its acceleration becomes 10 m/s2. The coefficient of friction is  (g = 10 m/s2)  

    Solution

  • Question 30
    1 / -0

    What is the smallest radius of a circle at which a bicyclist can travel if his speed is 7 m/s and the coefficient of static friction between tyres and road is 0.25  

    Solution

    Centripetal force = Frictional force
    ⇒ mv2/r​=μmg
    ∴r=v2​/μg
     
    ∴r=72​/0.25×10
    ∴r=19.6 m≈20 m

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