Work Energy and Power Test

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Work Energy and Power Test - 63

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Weekly Quiz Competition

Question 1
1 / -0

A block of mass $$10kg$$ is released on a fixed wedge inside a cart which is moved with constant velocity $$10 \ ms^{-1}$$ towards right. There is no relative motion between block and cart. Then work done by normal reaction on block in two seconds from ground frame will be $$(g=10 \ ms^{-2})$$:

A
$$1320 J$$

B
$$960 J$$

C
$$1200 J$$

D
$$240 J$$

Solution

Constant velocity means net force $$=0$$
Using Lami's theorem in the figure,
We have,

$$\displaystyle \dfrac {N}{sin (180^o- 53^o)}=\frac {100}{sin 90^o}$$

$$\therefore N=100 sin 53^o=80 N$$

Now, $$W_N=NS cos 53^o$$

$$=(80)(20)(0.6)$$

$$=960 J$$
Question 2
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A spring of natural length $$l$$ is compressed vertically downward against the floor so that its compressed length becomes $$\dfrac {l}{2}$$. On releasing, the spring attains its natural length. If $$k$$ is the stiffness constant of spring, then the work done by the spring on the floor is:

A
Zero

B
$$\dfrac {1}{2}kl^2$$

C
$$\dfrac {1}{2}l(\dfrac {1}{2})^2$$

D
$$kl^2$$

Solution

Only the center of mass of the spring changes, but the floor has no displacement.
Hence the work done by the spring on the floor is zero.
Option A.
Question 3
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A spring of spring constant $$5\times 10^3$$ N/m is stretched initially by $$5$$ cm from the unstretched position. The work required to further stretch the spring by another $$5$$ cm is:

A
6.25 N-m

B
12.50 N-m

C
18.75 N-m

D
25.00 N-m

Solution

Energy stored in a spring when stretched by $$x$$ is $$\dfrac {1}{2}kx^2$$

The amount of work required is the change in energy of the spring.
$$W=U_f-U_i=\dfrac {1}{2}k({x_2}^2-{x_1}^2)$$

$$\Rightarrow W=0.5(5000)(0.1^2-0.05^2)=18.75 \ N-m$$
Option C.
Question 4
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A spring of force constant $$k$$ is cut in two parts at its one-third length. When both the parts have same elongation, the work done in the two parts will be (Spring constant of a spring is inversely proportional to length of spring)

A
Equal to both

B
Greater for the longer part

C
Greater for the shorter part

D
Data insufficient

Solution

$$k\alpha \dfrac {1}{l}$$
l of shorter part is less, therefore value of k is more
$$W=\dfrac {1}{2}kx^2$$
$$\therefore W_{\text {shorter part}}$$ will be more.
Question 5
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A pendulum was kept horizontal and released. Find the acceleration of the pendulum when it makes an angle $$ \theta $$ with the vertical.

A
$$ g\ \sqrt { 1+3\cos ^{ 2 }{ \theta } } $$

B
$$ g\ \sqrt { 1+3\sin ^{ 2 }{ \theta } } $$

C
$$ g\ \sin { \theta } $$

D
$$ 2g\ \cos { \theta } $$

Solution

$$ \dfrac { { mv }^{ 2 } }{ 2 } $$ = $$ mgl\cos { \theta } $$ or
$$\dfrac { { v }^{ 2 } }{ l } $$ = $$ 2g\ \cos { \theta } $$
and $$ \sqrt { { a }_{ t }^{ 2 }+{ a }_{ r }^{ 2 } } =\sqrt { { \left( g\sin { \theta } \right) }^{ 2 }+{ (2g\quad \cos { \theta } ) }^{ 2 } } $$
= $$ g \ \sqrt { 1+3\cos ^{ 2 }{ \theta } } $$
Question 6
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A mass $${ m }_{ 1 }$$ with initial speed $${ v }_{ 0 }$$ in the positive $$x$$-direction collides with a mass $${ m }_{ 2 }=2{ m }_{ 1 }$$ which is initially at rest at the origin, as shown in figure. After the collision $${ m }_{ 1 }$$ moves off with speed $${ v }_{ 1 }={ v }_{ 0 }/2$$ in the negative $$y$$- direction, and $${ m }_{ 2 }$$ moves off with speed $${ v }_{ 2 }$$ at angle $$\theta$$. Find the velocity (magnitude and direction) of the centre of mass after the collision :

A
$${ v }_{ 0 }/3$$

B
$${ v }_{ 0 }/2$$

C
$${ v }_{ 0 }/5$$

D
$${ v }_{ 0 }/6$$

Solution

Since there is no external force acting on the 2 mass system, the momentum of the system and the velocity of the centre of mass remains constant.
Therefore, $$ V_{cm} $$ is the same before and after the collision, i.e;

$$ V_{cm,x} = \dfrac {m_1v_1+m_2v_2}{m_1
+m_2} $$, as $$m_2=2m_1$$ & $$ V_{cm,y}=0.$$

$$V_{cm,x}= \dfrac { m_1v_o+2m_1 \times 0}{m_1+2m_1} = \dfrac{v_o}{3} . $$
Question 7
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A body of mass $$m$$ is hauled from the Earth's surface by applying a force $$\vec{F}$$ varying with the height of ascent $$y$$ as $$\vec{F}=2(ay-1)mg$$, where $$a$$ is a positive constant. Find the work performed by this force $$W$$ and the increment in the body's potential energy $$\Delta U$$ in the gravitational field of the Earth over the first half of the ascent.

A
$$\displaystyle W=\frac{3mg}{4a}$$, $$\displaystyle\Delta U=\frac{mg}{2a}$$

B
$$\displaystyle W=\frac{3mg}{a}$$, $$\displaystyle\Delta U=\frac{mg}{2a}$$

C
$$\displaystyle W=\frac{3mg}{4a}$$, $$\displaystyle\Delta U=\frac{mg}{a}$$

D
$$\displaystyle W=\frac{3mg}{4a}$$, $$\displaystyle\Delta U=\frac{mg}{3a}$$

Solution

Force acting on the body of mass m=$$\vec { F } =2\left( a\times y-1 \right) \times m\times g$$
Body will be ascenting until $$\vec { F } =0\ \Rightarrow 2\left( a\times y-1 \right) \times m\times g=0\ \Rightarrow a\times y-1=0\ \Rightarrow a\times y=1\ \Rightarrow y =\dfrac { 1 }{ a }$$
So, the body will ascent until the height of y=$$\dfrac { 1 }{ a }$$
Now, the work done by the force $$\vec {F}$$ for half of the ascent height
$$=\int _{ 0 }^{ \dfrac { 1 }{ 2a } }{ \vec { F } } dy\ =\int _{ 0 }^{ \dfrac { 1 }{ 2a } }{ 2\left( a\times y-1 \right) } \times m\times g\times dy\ =\dfrac { 3mg }{ 4a }$$
Now, the potential energy stored in the body due to gravitational field force for half of ascent height $$=\dfrac{ mg }{ 2a } $$
Question 8
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A brick of mass $$1.8$$ kg is kept on a spring of spring constant $$K = 490 N m^{-1}$$. The spring is compressed so that after the release brick rises to $$3.6$$ m. Find the compression in the spring. (Take $$g= 10 \ m/s^2$$)

A
0.21 m

B
0.322 m

C
0.414 m

D
0.514 m

Solution

As the brick raises, it gains potential energy, which is the work done by the spring on the brick.

Let the spring be compressed by $$x$$, then the work done is $$\dfrac {1}{2}kx^2$$

$$\therefore \dfrac {1}{2}kx^2=mgh$$

$$\Rightarrow x=(\dfrac {2mgh}{k})^{1/2}=\sqrt {\dfrac {2\times 1.8\times 10\times 3.6}{490}}=0.514m$$

Option D.
Question 9
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A particle P of mass m attached to vertical axis by two strings AP and BP of length l each. The separation AB=l. The point p rotates around the axis with an angular velocity $$\omega $$. the tension in two strings are $$ { T }_{ 1 }$$ and $$ { T }_{ 2 } $$
taut only if $$ \omega >\sqrt { \frac { 2g }{ l } } $$

A
$$ { T }_{ 1 }$$= $$ { T }_{ 2 }$$

B
$$ { T }_{ 1 }$$+ $$ { T }_{ 2 }$$=$$\cfrac{2}{\sqrt3}m\omega^2l $$

C
$$ { T }_{ 1 }$$- $$ { T }_{ 2 }$$=2mg

D
BP will remain

Solution

From figure in equilibrium, $$T_1\cos60=mg+T_2\cos60$$ here $$\theta=60$$ for equilateral triangle.
or $$T_1(1/2)=mg+T_2(1/2) \Rightarrow T_1-T_2=2mg$$
also, $$T_1\sin60+T_2\sin60=m\omega^2l$$
or $$T_1+T_2=\frac{2}{\sqrt 3} m\omega^2l$$
Question 10
1 / -0

A spring of constant $$k$$ is fixed to a wall. A boy stretches this spring by distance $$x$$ and in the mean time the compartment moves by a distance $$s$$. The work done by boy with reference to earth is

A
$$\displaystyle \dfrac{1}{2} kx^2$$

B
$$\displaystyle \dfrac{1}{2} (kx)(s +x)$$

C
$$\displaystyle \dfrac{1}{2} kxs$$

D
$$\displaystyle \dfrac{1}{2} kx(s + x + s)$$

Solution

Work done by boy to stretch the spring
$$\displaystyle = \left ( \frac{1}{2} kx \right ) (S + x)$$
Work done by man on the floor
$$\displaystyle = - \left ( \frac{1}{2} kx \right ) (S)$$
$$\therefore$$ Total work done $$= \displaystyle \frac{1}{2} kx^2$$