Work Energy and Power Test

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Work Energy and Power Test - 80

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Weekly Quiz Competition

Question 1
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Two particles A and B having mass m each and charge $$q_1$$ and $$-q_2$$ respectively, are connected at the ends of a non conducting flexible and inextensible string of the length l. The particle A is fixed and B is whirled along a vertical circle with centre at A. If a vertically upward electric field of strength E exists in the space, then for minimum velocity of particle B:

A
The tension force in string at lowest point is zero

B
The tension force at the lowest point is non-zero

C
The tension force at the highest point is zero

D
The work done by interaction force between particles A and B is non-zero

Solution
Question 2
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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

A
$$20\ J$$

B
$$15\ J$$

C
$$10\ J$$

D
$$5\ J$$

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 3
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Two perfectly elastic objects $$A$$ and $$B$$ of identical mass are moving with velocities $$15\ m/s$$ and $$10\ m/s$$ respectively collide along the direction of line joining them. Their velocities after collision are respectively:

A
$$10\ m/s, 15\ m/s$$

B
$$20\ m/s, 5\ m/s$$

C
$$0\ m/s, 25\ m/s$$

D
$$5\ m/s, 20\ m/s$$

Solution

$$15m+10m=mv_1+mv_2$$
$$25=v_1+v_2$$...............(i)
and $$\dfrac{v_2-v_1}{u_1-u_2}=1$$
$$\Rightarrow \dfrac{v_2-v_1}{15-10}=1$$
$$\Rightarrow v_2-v_1=5$$............(ii)
$$v_1+v_2=25$$
$$\dfrac{v_2-v_1=5}{2v_2=30}$$
$$\therefore v_2=15m/s,v_1=10m/s$$
Question 4
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A heavy stone hanging from a massless string of length $$15$$m is projected horizontally with speed $$147$$ m/s. The speed of the particle at the point where the tension in the string equals the weight of the particle is?

A
$$10$$ m/s

B
$$7$$ m/s

C
$$12$$ m/s

D
None of these

Solution

Let at angle $$\theta$$ from vertical

speed be $$u$$ m/s and tension equal weight of particle i.e

$$\boxed{T=mg}$$.........(1) [given]

at this location, $$T-mg\cos\theta=\dfrac{mu^2}{l}$$

$$mg-mg\cos\theta=\dfrac{mu^2}{l}$$ [From (1)]

$$\boxed{lg(1-\cos\theta)=u^2}$$

by work energy theorem
$$W_{gravity}=\Delta K.E$$

$$-mgl(1-\cos\theta)=\dfrac{1}{2}mu^2-\dfrac{1}{2}mv^2$$

$$-2gl(1-\cos\theta)=u^2-v^2$$

$$-2u^2=u^2-v^2$$ [From (2)]

$$3u^2=v^2$$

$$u=\dfrac{v}{\sqrt{3}}=\dfrac{147}{\sqrt{3}}=85m/s$$

So answer is $$85m/s$$
Question 5
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A block of mass $$1\,kg$$ is free to move along the X-axis. It is at rest and from time $$t=0$$ onwards it is subjected to a time-dependent force $$F(t )$$ in the X-direction. The force $$F(t )$$ varies with $$t$$ as shown in figure. The kinetic energy of the block at $$t=4s$$ is

A
$$1J$$

B
$$2J$$

C
$$3J$$

D
$$0J$$

E
$$4J$$

Solution

$$\because$$ Total momentum of block = Area of given graph
$$\therefore P=A_1+A_2+A_3$$
Or $$P=\dfrac{1}{2}\times 1\times 2-\dfrac{1}{2}\times 2\times 2+\dfrac{1}{2}\times 1\times 2$$
Or $$P=0$$ or $$mv=0$$ or $$v=0$$
Hence, kinetic energy at $$(t=4s)=\dfrac{1}{2}mv^2=0$$
Question 6
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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

A
$$2(\sqrt {2} - 1) m/s$$

B
$$2(\sqrt {2} + 1) m/s$$

C
$$4.5\ m/s$$

D
None of these

Solution

According to question,
Initial kinetic energy, $$ E_1 = \dfrac{1}{2} mv^2 $$ ....(i)

Then, he increases his speed by $$2 \ m/s$$ 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
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Directions For Questions

A block of mass $$2\ m$$ collides elastically with a mass $$m$$ kept at rest. Friction exists between the block $$B$$ and surface with coefficient $$\mu =0.3$$, whereas no friction exists between block $$A$$ and the surface.

...view full instructions

After how much time collision between the blocks will not take place practically:

A
$$8\ sec$$

B
$$16\ sec$$

C
$$12\ sec$$

D
$$infinite$$

Solution
Question 8
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A body is falling a height $$h$$. After it has fallen a height $$\dfrac{h}{2}$$. it will possess

A
only potential energy

B
only kinetic energy

C
half potential energy and half kinetic energy

D
more kinetic and less potential energy

Solution

$$(c)$$ half potential and half kinetic energy
Explanation: When the body is height $$h$$; its potential energy is at maximum and kinetic energy is zero. When the body hits the ground, its potential energy becomes zero and kinetic energy is at maximum. At mid-way i.e., half the height; its potential energy becomes half of the maximum potential energy and same happens to the kinetic energy.
Question 9
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Four alternatives are given to each of the following incomplete statements/questions. Choose the right answer:
Which of the following object has higher potential energy?

A
mass $$ = 10\,kg$$ , $$ g = 9.8\, ms^{-2} , h = 10\,m $$

B
mass $$ = 5\,kg , g = 9.8\, ms^{-2} , h = 12\,m $$

C
mass $$ = 8\,kg , g = 9.8\,ms^{-2} , h = 100\,m $$

D
mass $$ = 6\,kg , g = 9.8\,ms^{-2} , h = 20\,m $$

Solution

The potential energy is given as-
$$PE = mgH$$
$$m$$; mass of the object
$$g = 9.8/ ms^{-2}$$
$$h$$; height gained from ground

For option A:
$$m = 10\ kg,\ h = 10\ m$$
$$PE = 10 \times 9.8 \times 10$$
$$\Rightarrow PE = 980\ J$$

For option B:
$$m = 5\ kg,\ h = 12\ m$$
$$PE = 5 \times 9.8 \times 12$$
$$\Rightarrow PE = 588\ J$$

For option C:
$$m = 8\ kg,\ h = 100\ m$$
$$PE = 8 \times 9.8 \times 100$$
$$\Rightarrow PE = 7840\ J$$

For option D:
$$m = 6\ kg,\ h = 20\ m$$
$$PE = 6 \times 9.8 \times 20$$
$$\Rightarrow PE = 1176\ J$$

For option C potential energy is highest so option C is the correct answer.
Question 10
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A force $$ F = K(yi + xj ) $$ ( where $$ K $$ is a positive constant) acts on a particle moving in the xy-plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0) and then parallel to the y-axis to the point (a, a). The total work done by the force F on the particles is

A
$$ -2 Ka^2 $$

B
$$ 2K a^2 $$

C
$$ -Ka^2 $$

D
$$ Ka^2 $$

Solution

While moving from$$ (0,0) to (a,0) $$
Along positive x-axis $$ y = 0 \therefore \overrightarrow {F} = -kx \hat {j} $$
i.e. force is in negative y-direction while displacement is in positive x-direction.
$$ \therefore W_1 = 0 $$
Because force is perpendicular to displacement
Then particle moves from $$ (a,0) $$ to $$ (a,a) $$ along a line parallel to y-axis $$ ( x = +a) $$ during this $$ \overrightarrow {F} = -k( y \hat {j} + a \hat {J}) $$
The first component of force $$ - ky \hat {i}$$ will not contribute any work because this component is along negative x-direction $$ ( - \hat {i}) $$ while displacement is in positive y- direction $$ (a,0) $$ to $$ (a,a) $$ The second component of force i.e. $$ - ka \hat {j} $$ will perform negative work.
$$ \therefore W_2 = ( - ka \hat {j})(a \hat {j}) = ( -ka)(a) = -ka^2 $$
So net work done on the particle $$ W = W_1 +W_2 $$
$$ = 0 + ( -ka^2) =-ka^2 $$