Work Energy and Power Test

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Work Energy and Power Test - 62

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Question 1
1 / -0

Calculate the energy possessed by a stone of mass $$10\ kg$$ kept at a height of $$5\ m$$.
Take $$g = 9.8\ m/s^{2}$$

A
$$196\ J $$

B
$$49\ J $$

C
$$490\ J $$

D
$$980\ J $$

Solution

The energy that stone may posses = kinetic energy + potential energy
Since, stone at rest $$(v = 0)$$
$$KE=\frac{mv^2}{2}$$ So kinetic energy=0
$$ \therefore $$ Total energy = Potential energy
$$(E) = mgh $$ [$$m$$ : mass , $$h$$ : height , $$g$$ : gravity ]

$$m = 10\ kg, g = 9.8\ m/s^{2} $$
and $$h = 5\ m $$
$$E = mgh = 10 \times 9.8 \times 5 $$ Joules
$$ = 490\ J $$
Question 2
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The work done by the force $$\vec{F}=A(y^{2}\hat{i}+2x^{2}\hat{j})$$, where $$A$$ is a constant and $$x$$ & $$y$$ are in meters around the path shown is :

A
$$zero$$

B
$$Ad$$

C
$$Ad^{2}$$

D
$$Ad^{3}$$

Solution

Variable force is apllied across the path $$OA$$ to $$AB$$ to $$BC$$ to $$CO$$

$$W=\int \bar { F } .d\bar { r }$$

$$ \\ =\int A(y^{ 2 }\bar { i } +2x^{ 2 }\bar { j } ).(dx\bar { i } +dy.\bar { j } )$$

$$\\ =\int A(y^{ 2 }dx+2x^{ 2 }dy)$$

Now following the path of displacement along $$OA$$ the $$y=0$$

$$\\ W_{ OA }=\int _{ x=0 }^{ x=d }{ A(0\quad +\quad 2 } x^{ 2 }.0)\quad =\quad 0+0\quad =0$$

Similerly, for $$W_{ AB },\quad x=d$$

$$\\ W_{ AB }=A[0+2d^{ 2 }d]\quad =\quad 2Ad^{ 3 }$$

Similarly, for $$W_{ BC },\ x=d,\quad y=d$$

$$\\ W_{ BC }=\int _{ x=d }^{ x=0 }{ A(d^{ 2 }dx\quad +\quad } 0)$$

$$\\ W_{ BC }=A[d^{ 2 }(-d)+0]=-Ad^{ 3 }$$

$$\\ W_{ CO }=A[0+0]$$

$$\\ W=0+2Ad^{ 3 }-Ad^{ 3 }+0=Ad^{ 3 }$$
Question 3
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A ball of mass m moves towards a moving wall of infinite mass with a speed 'v' along the normal to the wall. The speed of the wall is 'u' toward the ball. The speed of the ball after elastic collision with wall is

A
$$u + v$$ away from the wall

B
$$2u + v$$ away from the wall

C
$$|u - v|$$ away from the wall

D
$$|v- 2u |$$ away from the wall

Solution

In the frame of wall, the ball moves towards it with speed $$u + v$$ and due to momentum conservation velocity after collision returns with velocity $$u + v$$ in the direction in which wall is moving.
So from the ground frame, the velocity of ball after collision with wall is $$2u + v.$$
Question 4
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A force $$\vec{F}=(3t\hat{i}+5\hat{j})$$ N acts on a particle whose position vector varies as $$\vec{S}=(2t^{2}\hat{i}+5\hat{j})$$ m, where $$t$$ is time in seconds. The work done by this force from $$t=0$$ to $$t=2s$$ is:

A
23 J

B
32 J

C
zero

D
can't be obtained

Solution

The work done by the force $$\vec{F}=(3t\hat{i}+5\hat{j})$$ is

$$W=\int \vec{F}.d\vec{s}=\int (3t \hat{i}+5\hat{j}).(4t dt \hat{i})$$

$$=\int_{0}^{2}12t^{2}dt=\dfrac{12[t^{3}]_{0}^{2}}{3}=\dfrac{12(8-0)}{3}=32 \ J$$
Question 5
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A particle of mass $$m_0,$$ travelling at speed $$v_0,$$ strikes a stationary particle of mass $$2m_0.$$ As a result, the particle of mass $$m_0$$ is deflected through $$45^o$$ and has a final speed of $$\dfrac{v_0}{\sqrt2}$$. Then the speed of the particle of mass $$2m_0$$ after this collision is

A
$$\dfrac{v_0}{2}$$

B
$$\dfrac{v_0}{2\sqrt2}$$

C
$$\sqrt2v_0$$

D
$$\dfrac{v_0}{\sqrt2}$$

Solution

A particle of mass $$m_0,$$
Before collision

$$mv_{0}=2mv cos\theta+\displaystyle \dfrac{mv_{0}}{2}..........(i)$$

$$\displaystyle \theta=\dfrac{mv_{0}}{2}-2mv sin \theta............(ii)$$

By (i) & (ii), $$\theta=45^{0}$$
Now again momentum conservation in y direction

$$\displaystyle \dfrac{mv_{0}}{2}=2mv.\displaystyle \dfrac{1}{\sqrt2}\Rightarrow v=\displaystyle \dfrac{v_0}{2\sqrt2}$$
Question 6
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A particle of mass $$m$$ moves along the quarter section of the circular path whose centre is at the origin. The radius of the circular path is $$a$$. A force $$\vec{F}=y\hat{i}-x\hat{j}$$ newton acts on the particle, where $$x, y$$ denote the coordinates of position of the particle. The work done by this force in taking the particle from point A ($$a, 0$$) to point B ($$0, a$$) along the circular path is

A
$$\dfrac{\pi a^{2}}{4}J$$

B
$$\dfrac{\pi a^{2}}{2}J$$

C
$$\dfrac{\pi a^{2}}{3}J$$

D
$$None\ of\ these$$

Solution

$$\displaystyle W=\mathbf{\int F.dr}=\int (y\hat{i}-x\hat{j}).(dx\hat{i}+dy\hat{j})$$

$$=\displaystyle \int (ydx-xdy)$$ .........(1)

$$\because x^{2}+y^{2}=a^{2} \therefore xdx+y dy=0$$

$$\displaystyle \Rightarrow W=\int \left ( y\left ( \frac{-ydy}{x} \right )-xdy \right )=-\int \dfrac{x^{2}+y^{2}}{x}dy$$

$$\displaystyle =-\int_{0}^{a}\dfrac{a^{2}}{\sqrt{a^{2}-y^{2}}}dy=-\dfrac{\pi a^{2}}{2}$$

Alternate Method :
It can be observed that the force is tangent to the curve at
each point and the magnitude is constant. The direction of
force is opposite to the direction of motion of the particle.
$$\therefore$$ work done = (force) (distance)
$$=-\sqrt{x^{2}+y^{2}}=\dfrac{\pi a}{2}=-a\times \dfrac{\pi a}{2}=-\dfrac{\pi a^{2}}{2}J$$
$$W=-\dfrac{\pi a^{2}}{2}J$$
Question 7
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The force exerted by a compression device is given by $$ F(x) = kx (x-l)$$ for $$ 0 \leq x \leq l$$, where $$l$$ is the maximum possible compression, $$x$$ is the compression and $$k$$ is the constant. Work done to compress the device by a distance $$d$$ will be maximum when

A
$$d = \dfrac {l}{4}$$

B
$$d = \dfrac {l}{\sqrt{2}}$$

C
$$d = \dfrac {l}{2}$$

D
$$d = l $$

Solution

For W to be maximum ; $$\dfrac{dW}{dx} = 0$$;

i.e. $$ F(x) = 0 \Rightarrow x= l, x = 0$$

clearly for $$d = l $$, the work done is maximum.

Alternate Solution :
External force and displacement are in the same direction
$$\therefore$$ Work will be positive continuously so it will be maximum when displacement is maximum.
Question 8
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An elastic string carrying a body of mass $$m$$ at one end extends by $$1.5\ cm$$. If the body rotates in vertical circle with critical velocity, the extension in the string at the lowest position is:

A
$$3.0\ cm$$

B
$$4.5\ cm$$

C
$$1.5\ cm$$

D
$$9.0\ cm$$

Solution

When the body is at rest, the extension is $$1.5\ cm$$ while force acting on it is $$ mg $$
Force when it reaches bottom is $$F_b = \dfrac{mv^2}{r} + mg $$
Now using energy balance between topmost point and bottom most point ,
$$\dfrac{1}{2}mv^2=\dfrac{1}{2}m(\sqrt{rg})^2+mgh$$ (h is the height difference between the top and bottom point ; $$\sqrt{gr}$$ is critical velocity at topmost point )
$$ v^2 = rg + 2gh$$
or, $$ v^2 = rg + 4gr = 5 gr$$
So, $$F_b = 6mg $$
Using this in: $$\dfrac{e_2}{e_1} = \dfrac{F_2}{F_1} $$, we get:
$$e_2 = 1.5 \times 6 = 9.0\: cm $$
Question 9
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A force $$\vec F=(3t\hat i+5\hat j)$$N acts on a body due to which its displacement varies as $$\vec S=(2t^2\hat i-5\hat j)m$$. Work done by this force in $$2$$ second is

A
32 J

B
24 J

C
46 J

D
20 J

Solution

$$\vec v=\dfrac {d\vec S}{dt}=(4t)\hat i$$

$$P=\vec F\cdot \vec v=12 t^2$$

$$\therefore W=\int_0^2Pdt=\int_0^2 (12)t^2dt=32J$$.
Question 10
1 / -0

The world class Moses Mabhida football stadium situated in Durban has a symmetrical arc of length 350 m and a height of 106 m, as shown in the picture below on the left.
The picture on the right shows a funicular (Skycar), which takes tourists to the top of the arc. Suppose that the Skycar with tourists inside, starts from the base of the arc and travels a distance of 175 m along the arc to the viewing platform at the top. Assume that the work done by friction during the Skycars complete ascent is $$5.8\times 10^5J$$. If the combined mass of the Skycar and tourists is 5000 kg, then the work done by the motor that lifts the Skycar is approximately equal to,

A
$$4.6\times 10^6 J$$

B
$$5.8\times 10^6 J$$

C
$$8.0\times 10^6 J$$

D
$$9.2\times 10^6 J$$