Work Energy and Power Test

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Work Energy and Power Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

Kinetic energy of a body depends upon its:

A
mass

B
velocity

C
density

D
both A and B

Solution

Kinetic energy of the body $$K = \dfrac{1}{2}mv^2$$
$$\implies$$ $$K \propto m$$ and $$K\propto v^2$$
Thus kinetic energy of a body depends on the mass of the body as well as its velocity.
Question 2
1 / -0

A force of $$10 N$$ is applied on a object of mass $$1 kg$$ for $$2 s$$, which was initially at rest. What is the work done on the object by the force?

A
$$200 J$$

B
$$20 J$$

C
$$16 J$$

D
$$180 J$$

Solution

Force acting on the object $$F = 10 N$$
Mass of object $$m = 1kg$$
Acceleration of the object $$a = \dfrac{F}{m}$$
$$\therefore$$ $$a = \dfrac{10}{1} = 10 m/s^2$$
Initial speed of the object $$u = 0 m/s$$
Distance covered by it in $$2s$$, $$S = ut+\dfrac{1}{2}at^2$$ where $$t = 2s$$
$$\therefore$$ $$S = 0+\dfrac{1}{2}\times 10 \times 2^2 = 20 m$$
Work done $$W = FS $$
$$\implies$$ $$W = 10\times 20 = 200J$$
Question 3
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A $$500 g$$ particle tied to one end of a string is whirled in a vertical circle of circumference $$14 m$$.If the tension at the highest point of its path is $$2 N$$, what is its speed? (in m/s)

A
$$4.365$$

B
$$5.369$$

C
$$5.546$$

D
$$4.331$$

Solution

as the diagram shows
balancing the force at point $$P$$
given that $$T=2N$$, $$m=0.5kg$$ and $$R=\dfrac { 14 }{ 2\pi }=2.23m $$
$$\Rightarrow T+mg=m\dfrac { { v }^{ 2 } }{ R } $$
$$\Rightarrow v=\sqrt { R\dfrac { (T+mg) }{ m } } $$
$$\Rightarrow v=\sqrt { 2.23\times \dfrac { (2+0.5\times 10) }{ 0.5 } } =5.546m/s$$
Question 4
1 / -0

A force $$F_{x}$$ acts on a particle such that its position $$x$$ changes as shown in figure.
The work done by the particle as it moves from $$x = 0$$ to $$20\ m$$ is

A
$$37.5\ J$$

B
$$10\ J$$

C
$$15\ J$$

D
$$22.5\ J$$

E
$$45\ J$$

Solution

Key Concept As, we know, work done by a variable force, $$W_{n_{i}\rightarrow n_{t}}$$
$$=$$ area under the force $$-$$ displacement curve.
Work $$=$$ Area of $$\triangle ABF +$$ Area of $$\triangle BCEF +$$ Area of $$\triangle ECD$$
Work done by the particle as it moves from $$n = 0$$ to $$20\ m$$,
$$W = \dfrac {1}{2} \times 3\times 5 + 10\times 3 + \dfrac {1}{2} \times 5\times 3$$
$$\Rightarrow W = \dfrac {15}{2} + 30 + \dfrac {15}{2}$$
$$\Rightarrow W = 15 + 30 = 45\ J$$.
Question 5
1 / -0

A particle moves along X-axis from $$x = 10$$ to $$x = 5\ cm$$ under the influence of force given by $$F = (7 - 2x + 3x^{2}) N$$. The work done in the process is

A
$$70\ J$$

B
$$270\ J$$

C
$$35\ J$$

D
$$135\ J$$

Solution

The given force varies with distance.
Therefore, work done,
$$W = \displaystyle \int_{0}^{x = 5} F\cdot dx$$
$$W= \displaystyle \int_{0}^{5} (7 - 2x + 3x^{2})dx$$
$$W= \left [7x - \dfrac {2x^{2}}{2} + \dfrac {3x^{3}}{3} \right ]_{0}^{5}$$
$$W= [7 \times 5 - (5)^{2} + (5)^{3} - 0]$$
$$W= 135\ J$$.
Question 6
1 / -0

A stone of mass m is tied to a string and is moved in a vertical circle of radius r making n revolution per minute. The total tension in the string when the stone is at its lowest point is

A
$$mg$$

B
$$m(g+\pi n r^2)$$

C
$$m(g+ n r)$$

D
$$m(g+ \frac{\pi^2 n^2 r} {900})$$

Solution

Tension in string when it reach lower-most point
$$\displaystyle T=mg+m \omega^2 r$$
$$\displaystyle =m(g+4 \pi^2 n^2 r)$$ [as $$\omega=2 \pi n$$]
$$\displaystyle = m \left( g + 4 \pi^2 \left( \frac{n}{60} \right)^2 r \right)$$
$$\displaystyle = m \left( g + \left( \frac{\pi^2 n^2 r}{900} \right) \right)$$
Question 7
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A particle is moving in a vertical circle. The tension in the string when passing through two platform at angles $$\displaystyle 30^o$$ and $$\displaystyle 60^o$$ vertical (lowest position) are $$T_1$$ and $$T_2$$ respectively

A
$$\displaystyle T_1 = T_2$$

B
$$\displaystyle T_2 > T_1$$

C
$$\displaystyle T_1 > T_2$$

D
Tension in the string always remain the same

Solution

Tension, $$\displaystyle T= \frac{mv^2}{r} + mg cos \theta$$
For $$\displaystyle \theta =30^o$$
$$\displaystyle T_1= \frac{mv^2}{r} + mg cos 30^o$$
$$\displaystyle = \frac{mv^2}{r} + \frac{\sqrt{3}}{2} mg$$
For $$\displaystyle \theta =60^o$$
$$\displaystyle T_2= \frac{mv^2}{r} + mg cos 60^o$$
$$\therefore$$ $$\displaystyle \frac{mv^2}{r} + \frac{1}{2} mg$$
$$T_1> T_2$$
Question 8
1 / -0

A weightless thread can bear tension up to $$3.7$$kg wt. A stone of mass $$500$$g is tied to it and revolved in a circular path of radius $$4$$m in a vertical plane. If $$g=10m/s^2$$, then the maximum angular velocity of the stone will be :

A
$$4rad/s$$

B
$$2rad/s$$

C
$$6rad/s$$

D
none of these

Solution

Maximum tension $$=\displaystyle\frac{mv^2}{r}+mg$$
$$3.7\times 10=\displaystyle\frac{0.5v^2}{4}+0.5\times 10$$
$$\Rightarrow v=16m/s$$
$$\therefore \omega =\displaystyle\frac{v}{r}=\frac{16}{4}=4rad/s$$.
Question 9
1 / -0

Two wires are stretched through same distance. The force constant of second wire is half as that of the first wire. The ratio of work done to stretch first wire and second wire will be

A
$$2 : 1$$

B
$$1 : 2$$

C
$$3 : 1$$

D
$$1 : 3$$

Solution

As $$W=\dfrac { 1 }{ 2 } k{ x }^{ 2 }$$
If both wires are stretched through same distance, then $$W\propto k$$
It is given that $${ k }_{ 2 }=\dfrac { { k }_{ 1 } }{ 2 }$$
$$ \therefore \dfrac { { W }_{ 1 } }{ { W }_{ 2 } } =\dfrac { { k }_{ 1 } }{ { k }_{ 2 } }$$
$$ =\dfrac { { k }_{ 1 } }{ \dfrac { { k }_{ 1 } }{ 2 } } =\dfrac { 2 }{ 1 } =2 : 1$$
Question 10
1 / -0

A body of mass m is moving in a circle of radius r with a constant speed $$v$$. If a force $$\dfrac{mv^2}{r}$$ is acting on the body towards the centre, then what will be the work done by this force in moving the body over half the circumference of the circle?

A
Zero

B
$$\dfrac{mv^2}{r^2}$$

C
$$\dfrac{mv^2}{r^2}\times \pi r$$

D
$$\dfrac{\pi r^2}{mv^2}$$

Solution

Work done by centripetal force is always zero. As the force is acting towards the centre so, the work done by this force in moving the body over half the circumference of the circle is zero.

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Re-Attempt Weekly Quiz Competition

Work Energy and Power Test - 38

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Work Energy and Power Test - 38

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SHARING IS CARING

Question 1

mass

velocity

density

both A and B

Solution

Question 2

$$200 J$$

$$20 J$$

$$16 J$$

$$180 J$$

Solution

Question 3

$$4.365$$

$$5.369$$

$$5.546$$

$$4.331$$

Solution

Question 4

$$37.5\ J$$

$$10\ J$$

$$15\ J$$

$$22.5\ J$$

$$45\ J$$

Solution

Question 5

$$70\ J$$

$$270\ J$$

$$35\ J$$

$$135\ J$$

Solution

Question 6

$$mg$$

$$m(g+\pi n r^2)$$

$$m(g+ n r)$$

$$m(g+ \frac{\pi^2 n^2 r} {900})$$

Solution

Question 7

$$\displaystyle T_1 = T_2$$

$$\displaystyle T_2 > T_1$$

$$\displaystyle T_1 > T_2$$

Tension in the string always remain the same

Solution

Question 8

$$4rad/s$$

$$2rad/s$$

$$6rad/s$$

none of these

Solution

Question 9

$$2 : 1$$

$$1 : 2$$

$$3 : 1$$

$$1 : 3$$

Solution

Question 10

Zero

$$\dfrac{mv^2}{r^2}$$

$$\dfrac{mv^2}{r^2}\times \pi r$$

$$\dfrac{\pi r^2}{mv^2}$$

Solution

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