Work Energy and Power Test

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Work Energy and Power Test - 68

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Weekly Quiz Competition

Question 1
1 / -0

The bob $$A$$ of a pendulum of mass $$m$$ released from horizontal to the vertical hits another bob $$B$$ of the same mass at rest on a table as shown in figure. If the length of the pendulum is $$1\ m$$, what is the speed with which bob $$B$$ starts moving. (Neglect the size of the bobs and assume the collision to be elastic) (Take $$g = 10\ ms^{-2})$$.

A
$$4.47\ ms^{-1}$$

B
$$5.47\ ms^{-1}$$

C
$$6.47\ ms^{-1}$$

D
$$3.47\ ms^{-1}$$

Solution

As the collision is elastic and two balls have the same mass, therefore ball $$A$$ transfers its entire momentum to the ball $$B$$ and ball $$A$$ does not rise at all. After a collision, ball $$A$$ comes to rest and ball $$B$$ moves with speed of ball $$A$$.
$$\therefore$$ The speed with which bob $$B$$ starts moving is
$$v = \sqrt {2gh}$$

$$ = \sqrt {2\times 10\ ms^{-2} \times 1m} = \sqrt {20}\ ms^{-1}$$

$$= 4.47\ ms^{-1}$$
Question 2
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A small block of mass m is pushed on a smooth track from position A with a velocity $$2\sqrt5$$ times the minimum velocity required to reach point D. The block will leave the contact with track at the point where normal force between them becomes zero.
When the block reaches point B, what is the direction (in terms of angle with horizontal) of acceleration of the block?

A
$$tan^{-1}(\frac{1}{2})$$

B
$$tan^{-1}(2)$$

C
$$sin^{-1}(\frac{2}{3})$$

D
The block never reaches point B.

Solution
Question 3
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Two bodies $$A$$ and $$B$$ have masses $$20\ kg$$ and $$5\ kg$$ respectively. If they acquire the same kinetic energy. Find the ratio of thier velocities.

A
$$\dfrac {1}{2}$$

B
$$2$$

C
$$\dfrac {2}{5}$$

D
$$\dfrac {5}{6}$$

Solution

Given:

$$m_{A} = 20 kg$$

$$m_{B} = 5 kg$$

Kinetic Energy of both bodies A and B are the same.

Thus,

$$\dfrac{1}{2} m_{A} V_{A}^{2} = \dfrac{1}{2} m_{B} V_{B}^{2} \\$$

$$\dfrac{V_{A}^{2}}{V_{B}^{2}} = \dfrac{m_{B}}{m_{A}} \\$$

$$\dfrac{V_{A}^{2}}{V_{B}^{2}} = \dfrac{5}{20}\\$$

$$\dfrac{V_{A}}{V_{B}} = \dfrac{1}{2}$$

So, option A is correct.
Question 4
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A spring lies along the x-axis attached to a wall at one end and a block at the other end. The block rests on a friction less surface at x = 0. A force of constant magnitude F is applied to the block that begins to compress the spring, until the block comes to a maximum displacement $$x_{max}$$.
During the displacement, which of the curves shown in the graph best represents the work done on the spring block system by the applied force?

A
1

B
2

C
3

D
4

Solution

As $$'F'$$ is a constant force
Work done by $$F=F(displacement)\\ \quad=Fx\\ \Rightarrow a$$ straight line graph
$$(1)$$
Question 5
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An elastic spring of unstretched length $$L$$ and force constant $$K$$, is stretched by a small length $$x$$. It is further stretched by another small length $$y$$. calculate the work done during the second stretching is:

A
$$\cfrac { ky }{ 2 } \left( x+2y \right) $$

B
$$\cfrac { k }{ 2 } \left( 2x+y \right) $$

C
$$ky(x+2y)$$

D
$$\cfrac { ky }{ 2 } \left( 2x+y \right) $$

Solution

Initial stretching of the spring is $$x$$.
Initial potential energy   $$U_i = \dfrac{1}{2}kx^2$$
Final stretching of the spring is $$x+y$$.
Final potential energy of the spring   $$U_f = \dfrac{1}{2}k(x+y)^2 = $$ $$\dfrac{1}{2}kx^2 +\dfrac{1}{2}ky^2+\dfrac{1}{2}k(2xy)$$
Work done   $$W = U_f-U_i = $$ $$\dfrac{1}{2}kx^2 +\dfrac{1}{2}ky^2+\dfrac{1}{2}k(2xy) - \dfrac{1}{2}kx^2$$
$$\implies \ W = \dfrac{ky}{2}(2x+y)$$
Question 6
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A string wraps around a fat pipe as a bob attached to the string is made to move in a circular path in the horizontal. Assuming the speed is somehow held constant as the radius diminishes due to the wrapping, how will the centripetal force change?

A
It will stay the same.

B
It will diminish.

C
It will increase

D
None of the above (this is a jokeits got to be one of the three above)

Solution

As the string wraps, its radius decreases. Thus the centripetal force increases.

The correct option is (c)
Question 7
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A body is acted upon by a force which is proportional to the distance covered. If distance covered is represented by s, then work done by the force will be proportional to.

A
s

B
$$s^2$$

C
$$\sqrt{s}$$

D
None of the above

Solution

Given that-

$$F\propto s$$

$$\implies F=ks$$ where $$k=$$ proportionality constant

Now, we know that, Work done $$W=F.s$$

$$\implies W=ks^2$$

$$\implies F\propto s^2$$

Answer-(B)
Question 8
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Under the action of a force, $$2kg$$ body moves such that its position $$x$$ as a function of time is given by $$x={t}^{3}/3$$ where $$x$$ is in meters and $$t$$ in seconds. The work done by the force in the first two seconds is:

A
$$1.6J$$

B
$$16J$$

C
$$160J$$

D
$$1600J$$

Solution
Question 9
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A uniform elastic string has length $$a_{1}$$ when tension is $$T_{1}$$ and length $$a_{2}$$ when tension is $$T_{2}$$. The amount of work done in stretching it from its natural length to a length $$(a_{1}+ a_{2})$$ is ?

A
$$\dfrac {1}{2}\dfrac {(a_{1}T_{1}+a_{2}T_{2})^{2}}{2(T_{1}-T_{2})(a_{1}- a_{2})}$$

B
$$\dfrac {1}{2}\dfrac {(a_{1}T_{1}-a_{2}T_{2})^{2}}{2(T_{1}-T_{2})(a_{1}- a_{2})}$$

C
$$\dfrac {1}{2}\dfrac {(T_{1}+T_{2})^{2}}{2(T_{1}-T_{2})(a_{1}- a_{2})}$$

D
$$\dfrac {1}{2}\dfrac {(T_{1}+ T_{2})}{2(T_{1}-T_{2})(a_{1}- a_{2})}$$
Question 10
1 / -0

The position $$x$$ of a particle moving along $$x-$$axis at time $$(t)$$ is given by the equation $$t=\sqrt x+2$$, where $$x$$ is in metres and $$t$$ in seconds. Find the work done by the force in first four seconds.

A
$$Zero$$

B
$$2\ J$$

C
$$4\ J$$

D
$$8\ J$$

Solution

$$x=(t-2)^2=t^2-4t+4$$
$$v=2t-4$$
$$a=2$$
F=m.a.x
m=1kg
F=$$2\times (4^2-4\times 4+4)=8J$$