Work Energy and Power Test

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Work Energy and Power Test - 42

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Weekly Quiz Competition

Question 1
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A 0.25kg ball attached to a 1.5 m rope moves with a constant speed of 15 m/s around a vertical circle. Calculate the tension force on the rope at the middle of the circle:

A
37.5 N

B
137.5 N

C
2.5 N

D
25 N

Solution

The centripetal force acts along the direction of the tension and hence $$T=mv^2/R = 0.25 \times 15^2/1.5=37.5 N$$
Question 2
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two blocks of masses $$m_1 = 2 kg$$ and $$m_2 = 4 kg$$ are moving in the same direction with speeds $$\nu_1 = 6 m/s$$ and $$\nu_2 = 3 m/s$$, respectively on a frictioneless surface as shown in the figure. An ideal spring with spring constant $$k = 30000 N/m$$ is attached to the back side of $$m_2$$. Then the maximum compression of the spring after collision will be:

A
$$0.06 m$$

B
$$0.04 m$$

C
$$0.02 m$$

D
none of these

Solution

$$m_{1}=2\ kg ,m_{2}=4\ kg$$
$$v_{1}=6\ m/s, v_{2}=3\ m/s$$
$$k=30000\ N/m$$
From momentum conesrvation
$$2\times 6+5\times 3=2v_{1}+5v_{2}$$
$$2v_{1}+5v_{2}=27$$
$$v_{1}=v_{2}=v=\dfrac{27}{7}m/s$$ for max compression
From Energy conservation
$$\dfrac{1}{2}mv^{2}$$
$$\dfrac{1}{2}m_{1}v_{1}^{2}+\dfrac{1}{2}m_{2}v_{2}^{2}=\dfrac{1}{2}m_{2}v^{2}+\dfrac{1}{2}m_{2}v^{2}+\dfrac{1}{2}kx^{2}$$
$$\dfrac{1}{2}\times 2\times 6\times 6+\dfrac{1}{2}\times 4\times 3\times 3=\dfrac{1}{2}\times 2\times \dfrac{27\times 27}{7\times 7}+\dfrac{1}{2}\times \dfrac{4\times 27\times 27}{7\times 7}+\dfrac{1}{2}\times 30,000\times x^{2}$$
$$36+18=14.88+29.75+15000 x^{2}$$
$$54-44.63=15000 x^{2}\Rightarrow 937=15000 x^{2}$$
$$x=0.02\ m$$
Question 3
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A particle of mass 1 kg is suspended by means of a string of length L=2 m. The string makes $$6/\pi$$ rps around a vertical axis through the fixed end. The tension in the string is

A
72 N

B
36 N

C
288 N

D
10 N

Solution

The tension is the centripetal force = $$mw^2R$$

The angular velocity is w=$$(6/\pi)(2 \pi) =12$$ rads/s

Substituting we get, Tension = $$1 \times 144 \times 2 = 288 N$$

Thus the correct options is (c)
Question 4
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A mass attached to a string that is itself attached to the ceiling swings back and forth. If the bob is observed to be moving upward at a given instance, as shown to the right, which arrow best depicts the direction of the net force acting on the bob at that instant

A
A

B
B

C
C

D
D

Solution

Two forces act on the bob. (i) is the tension in the string and the (ii) mg sin $$\theta$$, which will be tangential to the path. The resultant of both the forces will be along vector C

Thus the correct option is (c)
Question 5
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When a rubber-bank is stretched by a distance $$x$$, it exerts a restoring force of magnitude $$F=ax+b{x}^{2}$$ where $$a$$ and $$b$$ are constants. The work done in stretching the unstretched rubber band by $$L$$ is:

A
$$\cfrac { a{ L }^{ 2 } }{ 2 } +\cfrac { b{ L }^{ 3 } }{ 3 } $$

B
$$\cfrac { 1 }{ 2 } \left( \cfrac { a{ L }^{ 2 } }{ 2 } +\cfrac { b{ L }^{ 3 } }{ 3 } \right) $$

C
$$a{ L }^{ 2 }+b{ L }^{ 3 }$$

D
$$\cfrac { 1 }{ 2 } \left( a{ L }^{ 2 }+b{ L }^{ 3 } \right) $$

Solution

$$F=ax+b{ x }^{ 2 }$$
$$dw=Fdx$$
$$W=\int _{ 0 }^{ L }{ \left( ax+b{ x }^{ 2 } \right) } dx\quad $$
$$W=\cfrac { a{ L }^{ 2 } }{ 2 } +\cfrac { b{ L }^{ 3 } }{ 3 } $$
Question 6
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A thin uniform rod of mass $$m$$ and length $$l$$ is hinged at the lower end of a level floor and stands vertically. It is now allowed to fall, then its upper and will strike the floor with a velocity given by(A)$$\sqrt { mgl }$$(B) $$\sqrt { 3gl }$$(c)$$\sqrt { 5gl }$$ (D) $$\sqrt { 2gl }$$ Sol.

A
A

B
B

C
C

D
D

Solution

Initially rod stand vertically

Its potential energy $$=mg=\dfrac{l}{2}$$

When it striker the floor, its potential energy will convert into rotational kinetic energy,

$$mg\left(\dfrac{1}{2}\right)=\dfrac{1}{2}I\omega^2$$

Where, $$I=\dfrac{ml^2}{3}=M.I$$ of rod about point A

$$\therefore mg\left(\dfrac{l}{2}\right)=\dfrac{l}{2}\left(\dfrac{ml^2}{3}\right)\left(\dfrac{V_B}{l}\right)^2$$

$$V_V=\sqrt{3gl}$$
Question 7
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A body of mass $$1kg$$ thrown upwards with a velocity of $$10m/s$$ comes to rest (momentarily) after moving up by $$4m$$. The work done by air drag in this process is (Take $$g=10m/{ s }^{ 2 }$$)

A
$$-20J$$

B
$$-10J$$

C
$$-30J$$

D
$$0J$$

Solution
Question 8
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A body is acted upon by force which is inversely proportional to the distance covered. The work done will be proportional to:

A
$$s$$

B
$${s}^{2}$$

C
$$\sqrt {s}$$

D
None of the above

Solution

Given $$F\propto \dfrac{1}{s} \implies F=\dfrac{k}{s}$$
$$W=\int _{ { s }_{ 1 } }^{ s }{ \overrightarrow { F } .d } \overrightarrow { s } =\int _{ { s }_{ 1 } }^{ s }{ \dfrac { k }{ s } .d } s=k\ln { \left( \dfrac{s}{s_1}\right) } $$
Question 9
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The work done by a force is equal to :

A
the area under $$f\left( x \right) $$ vs $$x$$ curve and $$x$$ axis

B
half the area under $$f\left( x \right) $$ vs $$x$$ curve and $$x$$ axis

C
the area under $$f\left( x \right) $$ vs $$x$$ curve and $$F$$ axis

D
half the area under $$f\left( x \right) $$ vs $$x$$ curve and $$F$$ axis

Solution

he correct option is A.

Therefore,

$$W=\int Fdx$$
Question 10
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A particle of mass $$2\ kg$$ travels along a straight line with velocity $$v=a \sqrt {X}$$, where $$a$$ is a constant. The work done by net force during the displacement of particle from $$x=0$$ to $$x=4\ m$$ is:

A
$$a^{2}$$

B
$$2a^{2}$$

C
$$4a^{2}$$

D
$$\sqrt {2} a^{2}$$

Solution

Given, $$mass=2kg, velocity=a\sqrt{x}$$

Since, $$v=a\sqrt{x}, v=a\sqrt4$$

So, $$V^2=u^2+2as\Rightarrow v^2-u^2=2as\Rightarrow v^2=2as\rightarrow a=\dfrac{v^2}{2s}$$ Since, $$u=0$$

$$a=\dfrac{2a^2}{2s}=\dfrac{4a^2}{4r^2}\Rightarrow a=\dfrac{a^2}{2}$$

So, workdone $$=f\times ds=2\times\dfrac{a^2}{2}\times 4=4a^2$$