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Work and Energy Test - 56

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Work and Energy Test - 56
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Weekly Quiz Competition
  • Question 1
    1 / -0
    A lift weighing $$250 Kg$$ is to be lifted up at a constant velocity of $$0.20 m/s$$. What would be the minimum horsepower of the motor to be used? 
    Solution

  • Question 2
    1 / -0
    The potential energy of a particle of mass 1 kg free to move along the x-axis is given by U($$x$$) =($$\frac{x^2}{2}$$-$$x$$) joule. If the total mechanical energy of the particle is 2J, then find the maximum speed of the particle. (Assuming only conservative force acts on the particle)
    Solution

  • Question 3
    1 / -0
    A ball is allowed to fall from a height of 10m. if there is $$40\%$$ loss of energy due is impact then after one impact ball will go up to
    Solution

  • Question 4
    1 / -0
    When a body falls freely towards the earth, then its total energy:
    Solution
    When a body falls freely towards the earth, the sum of the potential energy and kinetic energy of the object would be the same at all points. That is,
    $$potential\ energy + kinetic\ energy = constant$$
    It obeys the law of conservation of energy. Due to this, its total energy remains constant.
  • Question 5
    1 / -0
    A single conservative force acts on a $$1kg$$ particle  that moves along x-axis.Potential energy $$U(x)$$ is given by $$U(x)=20+{(x - 3)^2},$$ where x is in m.At $$x=0$$.particle has kinetic energy of $$20j$$ . 
    Find value of $$x$$ at which body is in equilibrium
    Solution

  • Question 6
    1 / -0
    When a man increases his speed by $$2 m/s$$, he finds that his kinetic energy is doubled, the original speed of the man is
    Solution
    According to question,
    Initial kinetic energy, $$ E_1 = \dfrac{1}{2} mv^2 $$         ....(i)

    Then, he increases his speed by $$2 \ m/s$$2m/s and K.E. is doubled.
    Final kinetic energy $$ E_2=2E_1= \dfrac{1}{2}m(v+2)^2 $$     ....(ii)

    Let multiply both sides of (i) by 2,
    $$2E_1=mv^2$$   . . . (iii)

    Equating (ii) and (iii),
    $$mv^2=\dfrac{1}{2}m(v+2)^2$$
    $$2v^2=v^2+4v+4$$
    Solving this using quadratic formula,
    $$ v=(2+2\sqrt{2})\ m/s $$
       $$= 2(1+\sqrt{2})\ m/s$$

    We reject the other value of $$v$$ as it must be positive because we considered $$v$$ to be speed which is a non-negative quantity or in this case, positive.

    Option B is correct.
  • Question 7
    1 / -0
    A constant force of $$5N$$ is applied on a block of mass $$20\ kg$$ for a distance of $$2.0\ m$$, the kinetic energy acquired by the block is
    Solution
    Acceleration $$a = \dfrac {F}{m} =\dfrac {5}{20} = \dfrac {1}{4} m/s^{2}$$
    $$ u =0 $$
    $$v = \sqrt {(2as)} = \sqrt {\left (\dfrac {2\times 2}{4}\right )} = 1\ m/s$$
    $$KE = \dfrac {1}{2} mv^{2} = \dfrac {1}{2} \times 20\times (1)^{2} = 10\ J$$

    (or)
     
    Work done, $$W = Fs$$ $$= 5\times 2 = 10\ J$$.
  • Question 8
    1 / -0
    A body at rest can have
    Solution
    A body at rest can have potential energy because PE is due to position of object but it does not have kinetic energy because KE is due to motion but object is at rest so no KE will be present.Momentum is also not present because momentum is due to velocity and velocity is zero in this case.
    For example an object at a roof at rest have no KE but it has gravitational potential energy due to it's height above ground given by $$mgh$$.
  • Question 9
    1 / -0
    The graph below show the force acting on a particle moves along the positive $$x-$$axis from the origin to $$x=x_1$$. The force if parallel to the $$x-$$axis and is conservative. The maximum magnitude $$F_1$$ has the same values for all graphs. Rank the solutions according to the change in the potential energy associated with the force. least (or most negative) to greatest (or most positive)  

    Solution

  • Question 10
    1 / -0
    A ball rolling on the ground possesses
    Solution
    A moving car possesses mechanical energy due to its motion (kinetic energy). A moving baseball possesses mechanical energy due to both its high speed (kinetic energy) and its vertical position above the ground (gravitational potential energy).
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