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Work Energy and Power Test - 35

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Work Energy and Power Test - 35
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  • Question 1
    1 / -0
    The water stored in a reservoir possesses :
    Solution
    Potential energy is the energy possessed due to the relative position of one body with respect to other.
    Kinetic energy is the energy possessed due to the motion of the body.
    Water stored in the reservoir is at rest, so no kinetic energy but it is at some height with respect to some level below the water surface, so it contains potential energy.
  • Question 2
    1 / -0
    An object of mass $$1  kg$$ has a P.E. of $$1  J$$ relative to the ground when it is at a height of
    Solution
    $$P.E$$ is given by $$=mgh$$
    where $$h$$ is the height, $$m$$ is mass and $$g$$ is acceleration due to gravity
    $$1J=1\times 10\times h\\ \\ $$
    $$h=0.1m$$

  • Question 3
    1 / -0
    The mass of an object $$P$$ is double the mass of $$Q$$. If both move with the same velocity, then the ratio of K.E. of $$P$$ to $$Q$$ is
    Solution
    $$ M_P =2 \times M_Q$$
    $$ V_P = V_Q $$

    $$ K.E_P: K.E_Q = ?$$

    $$\implies  \dfrac{1}{2}M_PV_P^2 : \dfrac{1}{2}M_QV_Q^2 $$
    Since $$V_P$$ =$$V_Q$$
    $$\implies M_P : M_Q $$
    $$\implies 2 \times M_Q : M_Q $$
    $$\implies 2 : 1 $$
  • Question 4
    1 / -0
    A boy of mass $$40\ kg$$ runs up a flight of $$50$$ steps, each $$10\ cm$$ high in $$14\ s$$. So, work done by the boy is:
    Solution
    Mass of the boy, $$m = 40\ kg$$
    Number of steps = $$50$$
    Height of each step = $$10\ cm$$
    Force on the boy due to gravity, $$F=mg=40\times{9.8}N=392\ N$$
    While climbing up the steps, the boy does work against gravity.
    Displacement in the vertical direction, $$s=(50\times{10})\ cm=500\ cm=5\ m$$
    Displacement is in the direction of force applied by the boy against gravity. 
    So, work done, $$W=F\times{s}=392\times{5}\ J=1960\ J$$
  • Question 5
    1 / -0
    Kinetic energy of a body depends on its:
    Solution
    $$K.E.=\dfrac{1}{2} mv^2$$ 

    So the kinetic energy of a body depends upon its mass and velocity.
  • Question 6
    1 / -0
    A fruit hanging from the top branch of a tree possesses
    Solution
    A fruit, hanging from the top branch of a tree, is at rest at a certain height from the earth"s surface. 
    So, it possesses gravitational potential energy.
  • Question 7
    1 / -0
    A stone tied to a string is rotated in a vertical plane. If mass of the stone is $$m$$, the length of the string is $$r$$ and the linear speed of the stone is $$v$$,When the stone is at its lowest point, then the tension in the string will be :
    Solution
    At the lowest point, as shown in the figure both mg and centrifugal force
    $$ \dfrac { m{ v }^{ 2 } }{ r } \\ $$will act in the same direction so,$$T=mg+\dfrac { m{ v }^{ 2 } }{ r } \\ $$
  • Question 8
    1 / -0
    In which form, the energy is stored in a fuel?
    Solution
    Chemical energy is energy stored in the bonds of atoms and molecules. Batteries, biomass, petroleum, natural gas, and coal are examples of chemical energy. Chemical energy is converted to thermal energy when people burn wood in a fireplace or burn gasoline in a car's engine.
  • Question 9
    1 / -0
    A string of length $$L$$ and force constant $$K$$ is stretched to obtain extension $$l$$. It is further stretched to obtain extension $${l}_{1}$$. The work done in second stretching is
    Solution
    Work done in stretching a string to obtain an extension $$l$$ is

    $${W}_{1} = \dfrac{1}{2}  Kl^2$$

    Similarly, work done in stretching a string to obtain extension $${l}_{1}$$ is

    $${W}_{2} = \dfrac{1}{2} K{l}_{1}^{2}$$

    $$\therefore$$     Work done in second case $$= {W}_{2} - {W}_{1} = \dfrac{1}{2} K \left({l}_{1}^{2} - {l}^{2}\right)$$
  • Question 10
    1 / -0
    A crane pulls up a car of mass $$500kg$$ to a vertical height of $$4m$$. So, work done by the crane is:
    Solution
    To raise the car, the crane has to do work against the force of gravity.
    Therefore, the force required to lift the car, $$F=mg=500\times{9.8}N=4900N$$
    Displacement, $$S=$$ vertical height raised $$=4m$$.
    $$\therefore$$ Work done, $$W=F.S=4900\times{4}J=19600J=19.6kJ$$
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