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Work Energy and Power Test - 15

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Work Energy and Power Test - 15
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  • Question 1
    1 / -0
    A ball tied to a string is swung in a vertical circle. Which of the following remains constant?
    Solution
    Earth's pull on the ball remains constant (W=mg)(W=mg).
    Tension in the string changes along the vertical circular path.
    Speed of the ball changes as sum of kinetic energy and potential energy remains constant during motion.
    Speed changes, hence, centripetal force changes.
  • Question 2
    1 / -0
    The K.E. of a body can be increased maximum by doubling its:
    Solution
    K.E=12mv2 K.E = \dfrac{1}{2}mv^2

    K.Em K.E \propto m     &    K.Ev2 K.E \propto v^2
    So doubling mass will double the kinetic energy and doubling speed will make kinetic energy 44 times.

    \therefore  The K.E. of a body can be increased maximum by doubling its speed.
  • Question 3
    1 / -0
    A lighter body moving with a velocity vv collides with a heavier body at rest. Then :
    Solution
    In a collision system where m1m_1 moves with u1u_1 initially and m2m_2 is at rest with m2>>>m1m_2 >>> m_1.
    Let the final velocity of m1m_1 be v1v_1 and m2m_2 be v2v_2.
    Assumption: Let the collision be elastic. Using linear momentum conservation and equation for coefficient of restitution,
    m1u1=m2v2+m1v1 m_1u_1 = m_2v_2 + m_1v_1
    v2v1=u1 v_2 - v_1 = u_1

    We get v1=m1m2m1+m2u1 v_1 = \dfrac {m_1 - m_2 }{m_1 + m_2} u_1

    v2=2m1m1+m2u1 v_2 = \dfrac {2m_1}{m_1 + m_2} u_1

    Using  m2>>> m1,m_2 >>>  m_1,  we get v1=u1 v_1 = -u_1 and v2=0 v_2 = 0 .  
  • Question 4
    1 / -0
    A certain force acting on a body of mass 2kg increase its velocity from 6m/s to 15 m/s in 2s. The work done by the force during this interval is ?
    Solution
    we know that
    v=u+at
    15=6+a(2)
    a=4.5m/s2a= 4.5 m/s^2
    s=ut+0.5at2=6(2)+0.5(4.5)(4)=21ms=ut+0.5at^2= 6(2)+ 0.5(4.5)(4)= 21 m
    W=mas=2(4.5)(21)=189JW= mas= 2(4.5)(21)= 189 J
  • Question 5
    1 / -0
    Name the type of energy (kinetic energy KK or potential energy UU) possessed in a compressed spring:
    Solution
    When you compress a spring, it possess potential energy. The force of compression is proportional to the compression, according to Hooke's Law. Releasing the spring turns the potential energy into kinetic energy. The spring can be then used to propel some object.
  • Question 6
    1 / -0
    Energy equals of a mass of one microgram in kilo joules is
    Solution
    Given
    Energy of 1 microgram=?
    Solution
    We will use Energy mass equivalence
    E=mc2E=mc^{2}
    E=106kg×(3×108)2m/s2E=10^{-6}kg\times (3 \times10^{8})^{2}m/s^{2}
    E=9×1010JE=9 \times 10^{10}J
    Option AA is  correct.
  • Question 7
    1 / -0
    A particle moves under the effect of a force F=CxF=Cx form x=0x=0 to x=x1x=x_{1}. The work done in the process is-
    Solution
    Similar to a spring with neutral point at x=0
    W=Fˉ.dˉs=0x1Cxdx=0.5Cx12W = \int \bar F.\bar ds = \int_0^{x_1} Cx dx = 0.5Cx_1^2
  • Question 8
    1 / -0
    A stone tied to a string is rotated in a vertical circle. The minimum speed with which the stone has to be rotated in order to complete the circle :
    Solution
    Step 1: Energy Conservation     [Ref. Fig.]\textbf{Step 1: Energy Conservation     [Ref. Fig.]}
    Dependence on m\text{Dependence on m}
    Let uu be the minimum speed required at BB.
    Apply energy conservation between point AA and BB we get
                  KEA+PEA=KEB+PEBKE_A + PE_A = KE_B + PE_B
    Taking point AA as zero potential level
        \Rightarrow            12mu2=mg(R+R)+12mv2\dfrac{1}{2} mu^2 = mg (R + R) + \dfrac{1}{2} mv^2   ....(1)....(1)

    From equation 11, we observe that mass mm is cancelled on both sides.
    u\Rightarrow u is independent of mm.

    Steo 2: Solving Equation\textbf{Steo 2: Solving Equation}
    Dependence on R\text{Dependence on R}
    At highest point, Tension will be zero for minimum velocity
    \Rightarrow gravitational pull will provide necessary centripetal force

               mg=mv2R\Rightarrow mg = \dfrac{mv^2}{R}

               v2=Rg\Rightarrow v^2 = Rg .... Putting in equation (1)(1)

    We get, 12mu2=mg(R+R)+12m(Rg)\dfrac{1}{2} mu^2 = mg (R + R) + \dfrac{1}{2} m (Rg)

               u=5Rg\Rightarrow u = \sqrt{5Rg}

    u\Rightarrow u is proportional to the square root of RR.
    Hence option BB is correct.

  • Question 9
    1 / -0
    Complete the following sentence:
    The kinetic energy of a body is the energy by virtue of its______ .
    Solution
    The kinetic energy of a body is the energy by virtue of its motion. This is the definition of kinetic energy. If it has motion then it will have velocity, hence kinetic energy exists.

    K.E=12mv2 K.E = \dfrac{1}{2}mv^2
  • Question 10
    1 / -0
    A simple pendulum, composed of a bob of mass mm connected to the end of a massless rod, executes simple harmonic motion as it swings through small angles of oscillation.The maximum angular displacement with the vertical is denoted by θ max{ \theta  }_{ max }. Frictional effects and variation of acceleration due to gravity are negligible.  
    Find out the correct statement?

    Solution
    We know the tangential acceleration in SHM is given by
    a=ω2Asinθa=-{\omega}^{2}A\sin{\theta}
        where A=amplitudeA=amplitude
    therefore at θ=0\theta=0
        a=ω2Asin00a=-{\omega}^{2}A\sin{0^0}
    or a=0a=0         because sin00=0\sin{0^0}=0
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