Self Studies

Work and Energy Test - 55

Result Self Studies

Work and Energy Test - 55
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    If water is flowing in a pipe at a height 4m from the ground then its potential
    energy per unit volume is (Reference is taken at ground, $$  g=10 \mathrm{m} / \mathrm{s}^{2} )  $$

  • Question 2
    1 / -0
    A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time $$t$$ is proportional to

    Solution

  • Question 3
    1 / -0
    If a body is released from a certain height, during its fall.
    Solution

  • Question 4
    1 / -0
    When the force and the displacement are in the opposite directions, the work done by the force is ..........
    Solution
    Work done by the force,
    $$W=\vec F.\vec r=Frcos\theta$$. . . . . . . .(1)
    If the force and the displacement are in the opposite direction, then angle between the force and the displacement is $$180^0$$
    $$cos\theta=cos180^0=-1$$
    from equation (1),
    $$W=-Fr$$
    The work done by the force is negative.
    The correct option is B.

  • Question 5
    1 / -0
    A machine delivering constant power moves a body along straight line starting from rest. The distance moved by the body in time $$t$$ is proportional to
    Solution
    Let the force applied on the body be $$F$$ and the distance travelled by it be $$s$$.
    We know that power $$P=\dfrac{work}{time}$$

                                        $$P=\dfrac{Fs}{t}=F \times \dfrac{s}{t}$$
    Speed $$v=\dfrac{Distance \ 's'}{Time \ 't'}$$

    So power $$P=Fv$$
    The power delivered by the machine is constant.
    $$\therefore P=Fv=constant \ 'k'$$ . . . (i)

    By Newton's second law $$Force \ 'F'=mass \ 'm' \times acceleration \ 'a'$$
    $$\Rightarrow F=ma$$  . . . (ii)

    By first equation of motion $$v=u+at$$
    $$\Rightarrow v=at$$ . . . (iii)
    ($$\because u=0$$ as body starts from rest)

    Using $$(ii)$$ and $$(iii)$$ in $$eq. (i)$$
    $$Fv=k$$
    $$(ma)(at)=k$$
    $$a^2=\dfrac{k}{mt}$$
    $$\Rightarrow a=\sqrt{\dfrac{k}{mt}}=\left( \dfrac{k}{mt}\right)^{\frac{1}{2}}$$ . . . (iv)

    From second equation of motion,
    $$s=ut+\dfrac{1}{2}at^2$$
    $$\Rightarrow s=\dfrac{1}{2}at^2$$ . . . (v)

    By substituting value of $$a$$ from $$(iv)$$ in $$(v)$$
    $$s=\dfrac{1}{2} \left( \dfrac{k}{mt}\right)^{\frac{1}{2}}t^2 $$

    $$s=\dfrac{1}{2} \left( \dfrac{k}{m}\right)^{\frac{1}{2}} \left( \dfrac{1}{t}\right)^\frac{1}{2} t^2 $$

    $$s=\left[ \dfrac{1}{2} \left( \dfrac{k}{m}\right)^{\frac{1}{2}} \right] (t)^{-1/2} (t)^2 $$

    The quantity in the square brackets $$[ \ ]$$ is constant so 
    $$s \propto (t)^{-1/2} (t)^2 $$

    $$s \propto t^{\frac{-1}{2}+2}$$

    $$s \propto t^{3/2}$$

    So option $$(B)$$ is the answer.
  • Question 6
    1 / -0
    A ball of mass 2 kg hits a floor with a speed of 4 m/s at an angle of $$ 60 ^o $$ with the normal.
    If (e=1/2); then the change in the kinetic energy of the ball is
    Solution

  • Question 7
    1 / -0
    From an automatic gun a man fires 360 bullets per minute with a speed of 360 km/hr. If each bullet weighs 20g, the power of the gun is
    Solution
    Power of gun  = $$ \dfrac { \text{Total K.E. of fired bullet} }{ \text{time }}  $$

    Total KInetic Energy $$=n\times \dfrac{1}{2}mv^2$$

     $$P = \dfrac {n \times \dfrac {1}{2}mv^2 }{t} = \dfrac {360}{60} \times  \dfrac { 1}{2} \times 2 \times 10^{-2} \times ( 100)^2 = 600 W $$

  • Question 8
    1 / -0
    A body of mass 200 g moving on a test has final K.E. of 50 J after travelling a distance of 10 cm. Assuming 90% loss of energy due to friction. The initial speed of the body is 
    Solution

  • Question 9
    1 / -0
    Potential energy v/s displacement curve for one dimensional conservative field is shown. Force at A and B is respectively -

    Solution

  • Question 10
    1 / -0
    A block of mass $$20 kg$$ is being brought down by a chain. If block acquires a speed of $$2 m/s$$ in dropping down $$2 m$$. Find work done by the chain during the process. ($$g= 10m/s^{2}$$ )
    Solution

Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Selfstudy
Selfstudy
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now