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Work Power Energy Test - 3
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Work Power Energy Test - 3

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  • Question 1/10
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    A particle is suspended vertically froma point O by an inextensible massless stringof length L. A vertical line AB is at a distance L/ 8 from O as shown in figure.
    The object is given a horizontal velocity u. At some point, its motion ceases to be circular and eventually the object passes through the line AB. At the instant of crossing AB, its velocity is horizontal. What will be the value of u?

    Solutions

    Let the string slacks at point Q as shown in figure. From P to Q path is circular and beyond Q path is parabolic. At point C, velocity of particle becomes horizontal, therefore, QD = half the range of the projectile.

    Now, we have following equations

    TQ = 0. Therefore, mg sin θ = mv2.....(i)

    v2 = u2 - 2gh = u2 - 2gL(1+ sin θ) .....(ii)


    QD = 1/2 (Range)

    Eq. (iii) can be written as

    Substituting this value of v2 in Eq. (ii)

    u2 = v2 + 2gL ( 1 + sinθ)

     

  • Question 2/10
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    Statement I A block of mass m starts moving on a rough horizontal surface with a velocity v. It stops due to friction between the block and the surface after moving through a certain distance. The surface is now tilted to an angle of 30° with the horizontal and the same block is made to go up on the surface with the same initial velocity v. The decrease in the mechanical energy in the second situation is smaller than that in the first situation.

    Statement II The coefficient of friction between the block and the surface decreases with the increase in the angle of inclination.

    Solutions

    In statement-I

    Decrease in mechanical energy in case I

    ΔU1 = 1/2 mv2

    But decrease in mechanical energy in case II will be

    ΔU2 = 1/2 mv2 - mgh

    ΔU2<ΔU1

    or statement I is correct.

    Statement IIis false, as the coefficient of friction is independent of the angle of inclination.

     

  • Question 3/10
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    A force   (where k is positive constant ) acts on a particle moving in the x - y plane; due to which the particle moves along the positive x - axis to the point (a,a), starting from origin. The total work done by the force F on the particle is

    Solutions

    While moving from (0, 0) to (a, 0) along positive x-axis,

    y = 0  

     i.e., force is in negative y-direction while the displacement is in positive x-direction. Therefore, W1 = 0 (Force ⊥L displacement).

    Then it moves from (a, 0) to (a, a) along a line parallel to y-axis (x = +a). During this 

    The first component of force, −ky will not contribute any work because this component is along negative x-direction while displacement is in positive y-direction (a,0) to (a, a).

     

    The second component of force i.e., −  will perform negative work 

    Note: In the given force, work done is path independent. It depends only on initial and final positions. Therefore, first method is brief and correct.

     

  • Question 4/10
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    A particle moving along x-axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as F (x) = - kx + ax3. Here, k and a are positive constants. For x ≥ 0 the functional from of the potential energy U (x) of the particle is

    Solutions

     

    Energy = work done on particle = −  over the range of displacement

    From the given function we can see that

    F = 0 at x = 0 i.e., slope of U - x graph is zero at x = 0.

    Therfore, this figure will correctly explain the change in potential energy.

     

  • Question 5/10
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    A simple pendulum is oscillating without damping. When the displacement of the bob is less then maximum, its acceleration vector  is correctly shown in

    Solutions

    Total acceleration consists of tangential acceleration due to gravity and the centripetal acceleration due to the circular motion of the bob , ar is towards centre and at is tangential hence sum in between it

    Net acceleration \vec{a} of the bob in position B has two components


    (a)  = radial acceleration (towards BA).
    (b)  = tangential acceleration (perpendicular to BA)

    Therefore, direction of   is correctly shown in option (c).

     

     

  • Question 6/10
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    List ParagraphAs given in the figures (a) and (b), AC, DG and GF are fixed inclined planes, BC = EF = x and AB = DE = y. A small block of mass M is released from the point A. It slides down AC and reaches C with a speed vC . The same block is released from rest from the point D. It slides down DGF and reaches the point F with speed vF. The coefficients of kinetic frictions between the block and both the surfaces AC and DGF are equal to μ. Calculate vC and vF.

    Solutions

    In both the cases work done by friction will be - μMgx. This is true for any type of inclination.

     

  • Question 7/10
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    Two blocks A and B are connected to each other by a string and a spring; the string passes over a frictionless pulley resting on another block C as shown in the figure.

    Block B slides over the horizontal top surface of a stationary block C and the block A slides along the vertical side of C, both with the same uniform speed. The coefficient of friction between the surfaces of blocks is 0.2. Force constant of the spring is 1960 N/m. If mass of block A is 2 kg. Calculate the mass of block B and the energy stored in the spring.

    Solutions

    Normal reaction between blocks A and C will be zero.Therefore, there will be no friction between them.
    Both A and B are moving with uniform speed. Therefore, net force on them should be zero.

    For equilibrium of A
    mA g = kx

     

  • Question 8/10
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    A string, with one end fixed on a rigid wall, passing over a fixed frictionless pulley at a distance of 2 m from the wall, has a mass M = 2kg attached to it at a distance of 1 m from the wall as shown in the figure. A mass m = 0.5 kg attached at the free end is held at rest so that the string is horizontal between the wall and the pulley and vertical beyond the pulley. Find the speed with which the mass M will hit the wall when the mass m is released. (Take g = 9.8 m/s2 )

    Solutions

    Let M strikes with speed v. Then, velocity of m at this instant will be v cosθ or  Further M will fall a distance of 1 m while m will rise up by m. From energy conservation :
    decrease in potential energy of M = increase in potential energy of m + increase in kinetic energy of both the blocks.

    Solving this equation, we get, v = 3.29 m/s.

     

  • Question 9/10
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    A spherical ball of mass m is kept at the highest point in the space between two fixed, concentric spheres A and B. The smaller sphere A has a radius R and the space between the two spheres has a width d. The ball has a diameter very slightly less than d. All surfaces are frictionless. The ball is given a gentle push (towards the right in the figure). The angle made by the radius vector of the ball with the upward vertical is denoted by θ.

    Which of the following expressions best describe the variation of the total normal reaction force exerted by the spheres on the ball as a function of angle θ.

    Solutions

    Velocity of ball at angleθ is

    Substituting value of v2 from Eq. (i), we get
    mg cos θ - N = 2 mg (1- cos θ)
    ∴N = mg (3 cos θ - 2)

     

  • Question 10/10
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    A block of mass 0.18 kg is attached to a spring of force constant 2 N/m. The coefficient of friction between the block and the floor is 0.1. Initially the block is at rest and the spring is unstretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. If initial velocity of the block in m/sis v = N/10,then N is equal to

    Solutions

    Decrease in mechanical energy = Work done against friction

    1/2 mv2 - 1/2 kx2 = μ mgx

     

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