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Work Energy and Power Test - 26

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Work Energy and Power Test - 26
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  • Question 1
    1 / -0
    A block is acted upon by a force, which is inversely proportional to the displacement xx. The work done will be proportional to
    Solution
    Given,  F=axF=\dfrac{a}{x} where a is a constant
    dW=Fds=axdxdW= \vec{F} \cdot \vec{ds}= \dfrac{a}{x}dx
    W=axdx=alnx\therefore W = \int_{}^{}\dfrac{a}{x}dx=a \ln x


  • Question 2
    1 / -0
    The hydroelectric plants do not produce electricity, if the water level in the dam is less than 34 m.
  • Question 3
    1 / -0
    A small sphere is attached to a cord and rotates in a vertical circle about a point OO. If the average speed of the sphere is increased, the cord is most likely to break at the orientation when the mass is at

    Solution
    Tension in the cord is maximum (for a given average speed of rotation) when the mass, mm, is at the bottom points B, as the gravitational force is in the downward direction and tension of the cord is directly opposing it.
  • Question 4
    1 / -0

    Directions For Questions

    A spring lies along xx axis attached to a wall at one end and a block at the other end. The block rests on a frictionless surface at x=0x=0.
    A force of constant magnitude FF is applied to the block that begins to compress the spring, until the block comes to a maximum displacement xmaxx_{max}.

    ...view full instructions

    During the displacement, which of the curves shown in the graph best represents the work done on the spring block system by the applied force ?

  • Question 5
    1 / -0
    By stretching the rubber strings of a catapult we store .......... energy in it.

    Solution
    Answer: Potential Energy
    The energy exerted or work done by our muscles on the rubber band is consumed in changing its shape. It gets stored in the stretched rubber band as its potential energy. It is this stored energy that is used by the rubber band to move to its original state, shape, and size. When the rubber band is released, this stored potential energy gets converted into kinetic energy.
    If a pebble is placed in contact with the stretched rubber band, this kinetic energy is transferred to the pebble. This kinetic energy of the pebble is enough to do some destructive work, like breaking a glass window, injuring someone, etc.
  • Question 6
    1 / -0
    A body starts from rest and acquires a velocity VV in time TT at constant rate. The work done on the body in time tt will be proportional to
    Solution
    Acceleration:  a=VTa = \dfrac{V}{T}

    Velocity at time t: v=at=VtTv=at=\dfrac{Vt}{T}

    Change in KE: ΔK=12mv20=12m(VtT)2=mV2t22T2\Delta K = \dfrac{1}{2}mv^2 -0 = \dfrac{1}{2}m(\dfrac{Vt}{T})^2= \dfrac{mV^2t^2}{2T^2}

    Work done on the body is equal to change in its kinetic energy.

    W=mV2t22T2 \therefore W = \dfrac{mV^2t^2}{2T^2}

    WV2t2T2 \Rightarrow W \propto \dfrac{V^2t^2}{T^2}
  • Question 7
    1 / -0
    A bob is suspended from a crane by a cable of length 5m5m. The crane and load are moving at a constant speed v0v_{0}. The crane is stopped by a bumper and the bob on the cable swings out an angle of 6060^{\circ}. Find the initial speed v0.(g=9.8m/s2)v_{0}.(g=9.8 m/s^{2})

    Solution
    By using conservation of energy, from the initial point to the highest point
    12mv02=mgl(1cos600)\frac { 1 }{ 2 } m{ { v }_{ 0 } }^{ 2 }=mgl(1-cos{ 60 }^{ 0 }) 
    v0=2gl(1cos600)=gl=50=7m/s\Rightarrow { v }_{ 0 }=\sqrt { 2gl(1-cos{ 60 }^{ 0 }) } =\sqrt { gl } =\sqrt { 50 } =7m/s
  • Question 8
    1 / -0
    A simple pendulum is vibrating with angular amplitude of θ=90o\theta=90^{o} as shown in figure.
    For what value of θ\theta is the acceleration directed
    (i)(i) Vertically upwards
    (ii)(ii) Horizontally
    (iii)(iii) Vertically downwards

    Solution
    Acceleration of a simple pendulum is a=ω2×A×cos (ωt+θ)a = -ω^{2}\times A\times cos  (ωt + θ), where ω is angular velocity, AA is amplitude, tt is time and θ is angle.
    If θ is 00, then a=ω2×A×cos(ωt)a = -ω^{2}\times A\times cos (ωt) so direction is vertically upwards, when θ is cos1(1/31/2)cos^{-1} (1/3^{1/2}) then direction is horizontally, when θ is 90o90^o, then a=ω2×A×sin(ωt)a = ω^{2}\times A\times sin (ωt) so direction is vertically downwards.
  • Question 9
    1 / -0
    A force F=(3ti^+5j^)N\displaystyle \vec{F}= \left ( 3t\hat{i}+5\hat{j} \right )N acts on a body due to which its displacement varies as S=(2t2i^5j^)m.\displaystyle \vec{S}= \left ( 2t^{2}\hat{i}-5\hat{j} \right )m. Work done by this force in 2 second is:
    Solution
    W=Fˉ.dsˉW = \int \bar F.\bar {ds}
    dˉs=4ti^dt\bar ds = 4t\hat i dt
    dW=(3ti^+5j^).(4ti^)dtdW = (3t\hat i + 5\hat j).(4t\hat i)dt
    W(t=2)=12t2dt=4t3=4(23)=32JW(t=2)=\int 12t^2dt = 4t^3 = 4(2^3) = 32J
  • Question 10
    1 / -0
    A stone tied to a string of length LL is swired in a verticle circle with the other end of the string at the centre. At a certain instant of time the stone is at it lowest position and has a speed uu. Find the magnitude of the change in its velocity as it reaches a position, where the string is horizontal.
    Solution
    By using conservation of energy,
    12mu2=mgL+12mv2\frac { 1 }{ 2 } m{ u }^{ 2 }=mgL+\frac { 1 }{ 2 } m{ v }^{ 2 }
    v=u22gL\Rightarrow v=\sqrt { { u }^{ 2 }-2gL }
    Change in velocity = u22gLi^uj^\sqrt { { u }^{ 2 }-2gL } \hat { i } -u\hat { j }
    Magnitude of change in velocity = u22gL+u2=2(u2gL) \sqrt { { u }^{ 2 }-2gL+{ u }^{ 2 } } =\sqrt { 2\left( { u }^{ 2 }-gL \right)  }

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