Work Energy and Power Test

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Work Energy and Power Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

A stone tied to a string of length $$L$$ is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time the stone is at its lowest position and has a speed $$u$$.The magnitude of change in its velocity as it reaches a position, where the string is horizontal is

A
$$\sqrt{u^2 - 2gL}$$

B
$$\sqrt{2gL}$$

C
$$\sqrt{u^2 - gL}$$

D
$$\sqrt{2(u^2 - gL)}$$

Solution

Here $$v^2 - u^2 = -2gl$$ ....(i)

$$v^2 = u^2-2gl$$

Since the velocities are mutually perpendicular, change in velocity

$$\triangle{v} = \sqrt{u^2 + v^2}$$ ....(ii)

$$=\sqrt{u^2 + u^2 - 2gl}$$

(substituting the value of $$v^2$$ from (i))

or $$\triangle{v} = \sqrt{2(u^2 - gl)}$$
Question 2
1 / -0

A stone tied to string of length $$l$$ is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time the stone is at its lowest position and has a speed $$u$$. The magnitude of the change in velocity as it reaches a position, where the string is horizontal is

A
$$\sqrt{u^2-2gl}$$

B
$$\sqrt{2gl}$$

C
$$\sqrt{u^2-gl}$$

D
$$\sqrt{2(u^2-gl)}$$
Question 3
1 / -0

A stone of mass $$1000g$$ tied to a light string of length $$10/3m$$ is whirling in a vertical circle. If the ratio of the maximum tension to minimum tension is $$4$$ and $$g=10{ ms }^{ -2 }$$, then the speed of stone at the highest point of circle is :

A
$$20{ ms }^{ -1 }$$

B
$$10\sqrt { 3 } { ms }^{ -1 }$$

C
$$5\sqrt { 3 } { ms }^{ -1 }$$

D
$$10{ ms }^{ -1 }$$

Solution

Minimum tension is at topmost point of the circle
Maximum tension is bottom point of the circle.
$$ \dfrac{Max. Tension}{Min. Tension} = \dfrac { \dfrac { m{ { v }_{ b }^{ 2 } } }{ r } +mg }{ \dfrac { m{ v }_{ t }^{ 2 } }{ r } -mg } =4 $$
and
Conservation of energy is :
$$ \dfrac { { v }_{ b }^{ 2 }-{ v }_{ t }^{ 2 } }{ 2 } =2gr $$
Using both equations,
$$ { v }_{ t }^{ 2 }=3gr$$
So
$$ {v}_{t} = \sqrt { 3gr } = 10\ m/s$$
Question 4
1 / -0

A particle of mass, $$m$$, is tied to a light string and rotated with a speed, $$v$$, along a circular path of radius, $$r$$. If $$T=$$ tension in the string and $$mg =$$ gravitational force on the particle, then the actual forces acting on the particle are

A
$$mg$$ and $$T$$ only.

B
$$mg$$, $$T$$ and an additional force of $$mv^2/r$$ directed inwards.

C
$$mg$$, $$T$$ and an additional force of $$mv^2/r$$ directed outwards.

D
Only a force $$mv^2/r$$ directed outwards.

Solution

The force $$mv^2/r$$ directed outwards, called centrifugal force, is not a real force.
At A, $$mv_1^2/l = T_1 +mg$$
The resultant of $$mg$$ and tension ($$T$$) will provide the centripetal force.
Question 5
1 / -0

A stone of mass $$1\ kg$$ tied to a light inextensible string of length. $$L = \dfrac{10}{3}$$ metre is whirling in a circular path of radius, $$L$$, in a vertical plane. If the ratio of maximum tension to the minimum tension is $$4$$ and if $$g$$ is taken to be $$10\ m/s^{2}$$, the speed of the stone at the highest point of circle is :

A
$$20\ m/s$$

B
$$10 \sqrt{3}\ m/s$$

C
$$10\ m/s$$

D
$$5\sqrt{2}\ m/s$$

Solution

When the stone is revolved in a vertical circle, tension is maximum at bottom and minimum at top. So we have, Tension at bottom as $$T_B$$ and Tension at top as $$T_T$$.

We know, $$\dfrac { { T }_{ B } }{ { T }_{ T } } =4$$

$$\Rightarrow \dfrac { \dfrac { mv_{ B }^{ 2 } }{ L } +mg }{ \dfrac { mv_{ T }^{ 2 } }{ L } -mg } =4$$

$$\Rightarrow \dfrac { mv_{ B }^{ 2 } }{ L } +mg=4\left( \dfrac { mv_{ T }^{ 2 } }{ L } -mg \right)$$

$$\Rightarrow \dfrac { mv_{ B }^{ 2 } }{ L } =\dfrac { 4mv_{ T }^{ 2 } }{ L } -5mg$$

$$\Rightarrow mv_{ B }^{ 2 } = 4mv_{ T }^{ 2 }-5mgL$$ ......$$(I)$$

Now using the law of conservation of energy, at top and bottom of the vertical circle, we have:

$$\dfrac { 1 }{ 2 } mv_{ T }^{ 2 }\quad +2mgL=\frac { 1 }{ 2 } mv_{ B }^{ 2 }$$

$$\Rightarrow m{ { v }_{ B } }^{ 2 }=m{ { v }_{ T } }^{ 2 }+4mgL$$ ......$$(II)$$

from $$(I)$$ and $$(II)$$, we have

$$4mv_{ T }^{ 2 }-5mgL=m{ { v }_{ T } }^{ 2 }+4mgL\\ \Rightarrow 3mv_{ T }^{ 2 }=9mgL\\ \Rightarrow { v }_{ T }=\sqrt { 3gL } \\ \Rightarrow { v }_{ T }=\sqrt { 3\times 10\times \left( \dfrac { 10 }{ 3 } \right) } =10\ m/s$$
Question 6
1 / -0

A body of mass $$m$$ is accelerated to velocity $$v$$ in time $$t'$$. The work done by the force as a function of time $$t$$ will be

A
$$\displaystyle \frac{m}{2} \frac{v^2 t^2}{t'^2}$$

B
$$\displaystyle \frac{1}{2} \left ( \frac{mv}{t'} \right )^2 t^2$$

C
$$\displaystyle \frac{m}{2} \frac{v}{t'} t^2$$

D
$$\displaystyle \frac{mvt^2}{2t'}$$

Solution

The force as a function of time $$t$$ on body of mass $$m$$ is
$$F = m.a$$ , $$ a$$-acceleration
And the work done by the force as a function of time $$t$$ is
$$W = F.d$$ , $$d$$-distance covered
$$W=ma.d$$
Here, the body is accelerated to velocity $$v$$ in time $$t'$$, hence
$$a = \dfrac{v}{t'}$$
distance traveled $$\displaystyle d=\int _{ 0 }^{ t }{ \frac { v }{ t' } tdt=\frac { v }{ t' } \frac { { t }^{ 2 } }{ 2 } } $$

$$\displaystyle W=m a. d = m \frac { {v}^{2} }{ {t'}^{2} } \frac { { t }^{ 2 } }{ 2 } $$
Question 7
1 / -0

The work done in stretching a spring of force constant $$K$$ from length $$l_1$$ to $$l_2$$ is :

A
$$K (l_2 - l_1)$$

B
$$\displaystyle \frac{K}{2} (l_2 + l_1)$$

C
$$K (l_2^2 - l_1^2)$$

D
$$\displaystyle \frac{K}{2} (l_2^2 - l_1^2)$$

Solution

Here, a spring of force constant K is stretched from length $$l_1$$ to $$l_2$$.
Let, the spring is stretched through a small distance $$x$$
The restoring force exerted by spring against the displacement x is
$$F' = - Kx$$
Hence, to stretch the spring the required force is equal and opposite to restoring force i.e.
$$F = -F' = Kx$$
Now, work done by applied force in stretching the spring is
$$\displaystyle W = \int_{{l}_{1}}^{{l}_{2}} F dx$$

$$\displaystyle W = \int_{{l}_{1}}^{{l}_{2}} Kx dx$$

$$\displaystyle W = \frac{K}{2}\left | x^2 \right |_{l_1}^{l_2}$$

$$\displaystyle W = \frac{K}{2}(l_2)^2 - (l_1)^2$$
Question 8
1 / -0

A locomotive of mass $$m$$ has a velocity $$v = a \sqrt{x}$$. Find the work done by all the forces acting on locomotive in first $$t$$ sec.

A
$$\displaystyle \frac{ma^2 t^2}{4}$$

B
$$\displaystyle \frac{ma^4 t^2}{4}$$

C
$$\displaystyle \frac{ma^4 t^2}{8}$$

D
$$\displaystyle \frac{ma^4 t^2}{2}$$

Solution

$$\displaystyle \dfrac{dv}{dt} = \dfrac{dv}{dx} \cdot \dfrac{dx}{dt} = \dfrac{a}{2 \sqrt{x}} (a \sqrt{x}) h = \dfrac{a^2}{2}$$
$$\therefore F = \displaystyle \dfrac{ma^2}{2}$$
$$\displaystyle S = u t + \dfrac{1}{2} Ft^2 = 0 + \dfrac{a^2}{4} t^2$$
$$W = \displaystyle F.s = \dfrac{ma^4 t^2}{8}$$
Question 9
1 / -0

Let $$\theta$$ denote the angular displacement of a simple pendulum oscillating in a vertical plane. If the mass of the bob is m, then tension in the string is mg $$\cos\theta$$

A
always

B
never

C
at the extreme position

D
at the mean position

Solution

For simple pendulum

$$\dfrac { m{ v }^{ 2 } }{ r } =T-mg\cos { \phi } $$

But when $$\phi =\theta $$, i.e., the bob is at extreme position. its velocity is zero, hence the equation becomes: $$T-mg\cos { \theta }=0$$

$$\Rightarrow \quad T = mg\cos { \theta } $$
Question 10
1 / -0

Water in a bucket is whirled in a vertical circle with a string to it. The water does not fallen even when the bucket is inverted at the top of its path . We conclude that in this position:

A
$$\displaystyle mg = \frac{mv^2}{r}$$

B
$$mg$$ is greater than $$\displaystyle \frac{mv^2}{r}$$

C
$$mg$$ is not greater than $$\displaystyle \frac{mv^2}{r}$$

D
$$mg$$ is not less than $$\displaystyle \frac{mv^2}{r}$$

Solution

Its because a centrifugal force $$\dfrac { m{ v }^{ 2 } }{ r } $$ acts on the water which is away from the center and opposite to force of gravity $$mg$$. If the $$mg$$ is greater then $$\dfrac { m{ v }^{ 2 } }{ r } $$ water will spill out. but if $$\dfrac { m{ v }^{ 2 } }{ r } $$ is greater then $$mg$$ water will not spill out of the bucket.

Note: for keeping the water in bucket while rotating it in a vertical circle you will have to keep the velocity above a certain value to make $$\dfrac { m{ v }^{ 2 } }{ r } $$ greater then $$mg$$.

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Work Energy and Power Test - 23

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Work Energy and Power Test - 23

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SHARING IS CARING

Question 1

$$\sqrt{u^2 - 2gL}$$

$$\sqrt{2gL}$$

$$\sqrt{u^2 - gL}$$

$$\sqrt{2(u^2 - gL)}$$

Solution

Question 2

$$\sqrt{u^2-2gl}$$

$$\sqrt{2gl}$$

$$\sqrt{u^2-gl}$$

$$\sqrt{2(u^2-gl)}$$

Question 3

$$20{ ms }^{ -1 }$$

$$10\sqrt { 3 } { ms }^{ -1 }$$

$$5\sqrt { 3 } { ms }^{ -1 }$$

$$10{ ms }^{ -1 }$$

Solution

Question 4

$$mg$$ and $$T$$ only.

$$mg$$, $$T$$ and an additional force of $$mv^2/r$$ directed inwards.

$$mg$$, $$T$$ and an additional force of $$mv^2/r$$ directed outwards.

Only a force $$mv^2/r$$ directed outwards.

Solution

Question 5

$$20\ m/s$$

$$10 \sqrt{3}\ m/s$$

$$10\ m/s$$

$$5\sqrt{2}\ m/s$$

Solution

Question 6

$$\displaystyle \frac{m}{2} \frac{v^2 t^2}{t'^2}$$

$$\displaystyle \frac{1}{2} \left ( \frac{mv}{t'} \right )^2 t^2$$

$$\displaystyle \frac{m}{2} \frac{v}{t'} t^2$$

$$\displaystyle \frac{mvt^2}{2t'}$$

Solution

Question 7

$$K (l_2 - l_1)$$

$$\displaystyle \frac{K}{2} (l_2 + l_1)$$

$$K (l_2^2 - l_1^2)$$

$$\displaystyle \frac{K}{2} (l_2^2 - l_1^2)$$

Solution

Question 8

$$\displaystyle \frac{ma^2 t^2}{4}$$

$$\displaystyle \frac{ma^4 t^2}{4}$$

$$\displaystyle \frac{ma^4 t^2}{8}$$

$$\displaystyle \frac{ma^4 t^2}{2}$$

Solution

Question 9

always

never

at the extreme position

at the mean position

Solution

Question 10

$$\displaystyle mg = \frac{mv^2}{r}$$

$$mg$$ is greater than $$\displaystyle \frac{mv^2}{r}$$

$$mg$$ is not greater than $$\displaystyle \frac{mv^2}{r}$$

$$mg$$ is not less than $$\displaystyle \frac{mv^2}{r}$$

Solution

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