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Work Energy and Power Test - 63

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Work Energy and Power Test - 63
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  • Question 1
    1 / -0
    A block of mass 10kg10kg is released on a fixed wedge inside a cart which is moved with constant velocity 10 ms110 \ ms^{-1} towards right. There is no relative motion between block and cart. Then work done by normal reaction on block in two seconds from ground frame will be (g=10 ms2)(g=10 \ ms^{-2}):

    Solution

    Constant velocity means net force =0=0
    Using Lami's theorem in the figure,
    We have,

    Nsin(180o53o)=100sin90o\displaystyle \dfrac {N}{sin (180^o- 53^o)}=\frac {100}{sin 90^o}

    N=100sin53o=80N\therefore N=100 sin 53^o=80 N

    Now, WN=NScos53oW_N=NS cos 53^o

    =(80)(20)(0.6)=(80)(20)(0.6)

    =960J=960 J

  • Question 2
    1 / -0
    A spring of natural length ll is compressed vertically downward against the floor so that its compressed length becomes l2\dfrac {l}{2}. On releasing, the spring attains its natural length. If kk is the stiffness constant of spring, then the work done by the spring on the floor is:
    Solution
    Only the center of mass of the spring changes, but the floor has no displacement.
    Hence the work done by the spring on the floor is zero.
    Option A.
  • Question 3
    1 / -0
    A spring of spring constant 5×1035\times 10^3 N/m is stretched initially by 55 cm from the unstretched position. The work required to further stretch the spring by another 55 cm is:
    Solution
    Energy stored in a spring when stretched by xx is 12kx2\dfrac {1}{2}kx^2

    The amount of work required is the change in energy of the spring.
    W=UfUi=12k(x22x12)W=U_f-U_i=\dfrac {1}{2}k({x_2}^2-{x_1}^2)

    W=0.5(5000)(0.120.052)=18.75 Nm\Rightarrow W=0.5(5000)(0.1^2-0.05^2)=18.75 \ N-m
    Option C.
  • Question 4
    1 / -0
    A spring of force constant kk is cut in two parts at its one-third length. When both the parts have same elongation, the work done in the two parts will be (Spring constant of a spring is inversely proportional to length of spring)
    Solution
    kα1lk\alpha \dfrac {1}{l}
    l of shorter part is less, therefore value of k is more
    W=12kx2W=\dfrac {1}{2}kx^2
    Wshorter part\therefore W_{\text {shorter part}} will be more.
  • Question 5
    1 / -0
    A pendulum was kept horizontal and released. Find the acceleration of the pendulum when it makes an angle θ \theta with the vertical.

    Solution
    mv22 \dfrac { { mv }^{ 2 } }{ 2 } = mglcosθ  mgl\cos { \theta  } or
    v2l\dfrac { { v }^{ 2 } }{ l } = 2g cosθ  2g\ \cos { \theta  }
    and at2+ar2=(gsinθ  ) 2+(2gcosθ )2 \sqrt { { a }_{ t }^{ 2 }+{ a }_{ r }^{ 2 } } =\sqrt { { \left( g\sin { \theta  }  \right)  }^{ 2 }+{ (2g\quad \cos { \theta  } ) }^{ 2 } }
    = g 1+3cos2θ   g \ \sqrt { 1+3\cos ^{ 2 }{ \theta  }  }  
  • Question 6
    1 / -0
    A mass m1{ m }_{ 1 } with initial speed v0{ v }_{ 0 } in the positive xx-direction collides with a mass m2=2m1{ m }_{ 2 }=2{ m }_{ 1 } which is initially at rest at the origin, as shown in figure. After the collision m1{ m }_{ 1 } moves off with speed v1=v0/2{ v }_{ 1 }={ v }_{ 0 }/2 in the negative yy- direction, and m2{ m }_{ 2 } moves off with speed v2{ v }_{ 2 } at angle θ\thetaFind the velocity (magnitude and direction) of the centre of mass after the collision :

    Solution
    Since there is no external force acting on the 2 mass system, the momentum of the system and the velocity of the centre of mass remains constant.
    Therefore, Vcm V_{cm} is the same before and after the collision, i.e;

    $$ V_{cm,x} = \dfrac {m_1v_1+m_2v_2}{m_1
    +m_2} ,as, as m_2=2m_1$$   &  $$ V_{cm,y}=0.$$

    Vcm,x=m1vo+2m1×0m1+2m1=vo3.V_{cm,x}= \dfrac { m_1v_o+2m_1 \times 0}{m_1+2m_1} = \dfrac{v_o}{3} .
  • Question 7
    1 / -0
    A body of mass mm is hauled from the Earth's surface by applying a force F\vec{F} varying with the height of ascent yy as F=2(ay1)mg\vec{F}=2(ay-1)mg, where aa is a positive constant. Find the work performed by this force WW and the increment in the body's potential energy ΔU\Delta U in the gravitational field of the Earth over the first half of the ascent.
    Solution
    Force acting on the body of mass m=F=2(a×y1)×m×g\vec { F } =2\left( a\times y-1 \right) \times m\times g
    Body will be ascenting until F=0 2(a×y1)×m×g=0 a×y1=0 a×y=1 y=1a\vec { F } =0\ \Rightarrow 2\left( a\times y-1 \right) \times m\times g=0\ \Rightarrow a\times y-1=0\ \Rightarrow a\times y=1\ \Rightarrow y =\dfrac { 1 }{ a }
    So, the body will ascent until the height of y=1a\dfrac { 1 }{ a }
    Now, the work done by the force F\vec {F}  for half of the ascent height
    =012a F dy =012a 2(a×y1) ×m×g×dy =3mg4a=\int _{ 0 }^{ \dfrac { 1 }{ 2a }  }{ \vec { F }  } dy\ =\int _{ 0 }^{ \dfrac { 1 }{ 2a }  }{ 2\left( a\times y-1 \right)  } \times m\times g\times dy\ =\dfrac { 3mg }{ 4a }
    Now, the potential energy stored in the body due to gravitational field force for half of ascent height =mg2a=\dfrac{ mg }{ 2a }
  • Question 8
    1 / -0
    A brick of mass 1.81.8 kg is kept on a spring of spring constant K=490 N m1K = 490  N  m^{-1}. The spring is compressed so that after the release brick rises to 3.63.6 m. Find the compression in the spring. (Take g=10 m/s2g= 10 \ m/s^2)
    Solution
    As the brick raises, it gains potential energy, which is the work done by the spring on the brick.

    Let the spring be compressed by xx, then the work done is 12kx2\dfrac {1}{2}kx^2

    12kx2=mgh\therefore \dfrac {1}{2}kx^2=mgh

    x=(2mghk)1/2=2×1.8×10×3.6490=0.514m\Rightarrow x=(\dfrac {2mgh}{k})^{1/2}=\sqrt {\dfrac {2\times 1.8\times 10\times 3.6}{490}}=0.514m

    Option D.
  • Question 9
    1 / -0
    A particle P of mass m attached to vertical axis by two strings AP and BP of length l each. The separation AB=l. The point p rotates around the axis with an angular velocity ω\omega . the tension in two strings are T1 { T }_{ 1 } and T2 { T }_{ 2 }
    taut only if ω>2gl  \omega >\sqrt { \frac { 2g }{ l }  }

    Solution
    From figure in equilibrium, T1cos60=mg+T2cos60T_1\cos60=mg+T_2\cos60   here θ=60\theta=60 for equilateral triangle.
    or T1(1/2)=mg+T2(1/2)T1T2=2mgT_1(1/2)=mg+T_2(1/2) \Rightarrow T_1-T_2=2mg
    also, T1sin60+T2sin60=mω2lT_1\sin60+T_2\sin60=m\omega^2l
    or T1+T2=23   mω2lT_1+T_2=\frac{2}{\sqrt 3}   m\omega^2l

  • Question 10
    1 / -0
    A spring of constant kk is fixed to a wall. A boy stretches this spring by distance xx and in the mean time the compartment moves by a distance ss. The work done by boy with reference to earth is

    Solution
    Work done by boy to stretch the spring
    = (12kx)(S+x)\displaystyle = \left ( \frac{1}{2} kx \right ) (S + x)
    Work done by man on the floor
    =(12kx)(S)\displaystyle = - \left ( \frac{1}{2} kx \right ) (S)
    \therefore Total work done =12kx2= \displaystyle \frac{1}{2} kx^2
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