Work Energy and Power Test

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Work Energy and Power Test - 65

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Question 1
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A simple pendulum is released from rest with the string in horizontal position. The vertical component of the velocity of the bob becomes maximum, when the string makes an angle $$\displaystyle \theta $$ with the vertical. The angle $$\displaystyle \theta $$ is equal to

A
$$\displaystyle \frac{\pi }{4}$$

B
$$\displaystyle \cos ^{-1}\left ( \frac{1}{\sqrt{3}} \right )$$

C
$$\displaystyle \sin ^{-1}\left ( \frac{1}{\sqrt{3}} \right )$$

D
$$\displaystyle \frac{\pi }{3}$$

Solution

At angle theta , l = length of rod v = pendulum velocity u = vertical velocity
By energy conservation,
$$mgl\cos { \theta } = \dfrac { m{ v }^{ 2 } }{ 2 } = \dfrac { m{ u }^{ 2 } }{ 2\sin ^{ 2 }{ \theta } } $$
Differentiating wrt $$\theta$$ , and putting $$\dfrac { du }{ d\theta } =0$$;
we get, $$\theta = \cos ^{ -1 }{ \left( \dfrac { 1 }{ \sqrt { 3 } } \right) } \\ $$
Question 2
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A force $$\displaystyle F= -\frac{k}{x^{2}}\left ( x\neq 0 \right )$$ acts on a particle in x-direction. Find the work done by this force in displacing the particle from. $$\displaystyle x= +a\:$$to$$\:x= +2a.$$ Here, $$k$$ is a positive constant.

A
$$\displaystyle -\frac{k}{2a}$$

B
$$\displaystyle \frac{k}{2a}$$

C
$$\displaystyle -\frac{k}{a}$$

D
$$\displaystyle +\frac{k}{a}$$

Solution

$$\displaystyle W= \int F\:dx= \int_{+a}^{+2a}\left ( \frac{-k}{x^{2}} \right )dx= \left [ \frac{k}{x} \right ]_{+a}^{+2a}$$
$$\displaystyle = -\frac{k}{2a}$$
Note: It is important to note that work comes out to be negative which is quite obvious as the force acting on the particle is in negative x-direction $$\displaystyle \left ( F= -\frac{k}{x^{2}} \right )$$ while displacement is along positive x-direction. (from x=a to x=2a)
Question 3
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An object is displaced from position vector $$\displaystyle \vec{r}_{1}= \left ( 2\hat{i}+3\hat{j} \right )m\:$$ to$$ \:\vec{r}_{2}= \left ( 4\hat{i}+6\hat{j} \right )$$ m under a force $$\displaystyle \vec{F}= \left ( 3x^{2}\hat{i}+2y\hat{j} \right )N.$$ Find the work done by this force.

A
$$83 J$$

B
$$41.5 J$$

C
$$166 J$$

D
$$164 J$$

Solution

$$\displaystyle W= \int_{\vec{r}_{1}}^{\vec{r}_{2}}\vec{F}\cdot \vec{dr}$$

$$\displaystyle = \int_{\vec{r}_{1}}^{\vec{r}_{2}}\left ( 3x^{2}\hat{i}+2y\hat{j} \right )\cdot \left ( dx\hat{i}+dy\hat{j}+dz\hat{k} \right )$$

$$\displaystyle = \int_{\vec{r}_{1}}^{\vec{r}_{2}}\left ( 3x^{2}dx+2ydy \right )= \left [ x^{3}+y^{2} \right ]_{\left ( 2,3 \right )}^{\left ( 4,6 \right )}$$

$$=83 J$$
Question 4
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A particle is moving in a circular path in the vertical plane. It is attached at one end of a string of length $$l$$ whose other end is fixed. The velocity at lowest point is $$u$$. The tension in the string is $$\displaystyle \vec{T}$$ and acceleration of the particle is $$\displaystyle \vec{a}$$ at any position. Then $$\displaystyle \vec{T}.\vec{a}$$ is zero at highest point if

A
$$\displaystyle u> \sqrt{5gl}$$

B
$$\displaystyle u= \sqrt{5gl}$$

C
Both (a) and (b) correct

D
Both (a) and (b) are wrong

Solution

Tension is $$0$$ in case of $$B$$.
If $$u$$ is more than this, then Tension is never zero.
if $$u$$ is less than this, then the particle would never be able to complete the circular loop.
$$m{ v }^{ 2 }/r\quad =\quad mg\\ \dfrac { m{ u }^{ 2 } }{ 2 } =\quad 2\times mgR\quad +\quad \dfrac { mv^{ 2 } }{ 2 } \\ \Rightarrow \quad u\quad =\quad \sqrt { 5Rg } $$
Question 5
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A cutting tool under microprocessor control has several forces acting on it. One force is $$\vec{F}=-\alpha xy^2\hat{j}$$, a force in the negative y-direction whose magnitude depend on the position of the tool. The constant is $$\alpha = 2.50\ N/m^3$$. Consider the displacement of the tool from the origin to the point $$x = 3.00 \,m, y = 3.00 \,m$$. Calculate the work done on the tool by $$\vec{F}$$ if the tool is first moved out along the x-axis to the point $$x = 3.00m, \:y= 0m $$ and then moved parallel to the y-axis to $$x = 3.00m, y = 3.00 \,m$$.

A
$$67.5 J$$

B
$$85 J$$

C
$$102 J$$

D
$$7.5 J$$

Solution

as , $$dW=f.ds$$

$$W=\int { \vec{F}.\vec{ds} } $$
and force is only along y-axis ,work done along x- axis is zero
work done in moving from $$(3,0)$$ to $$(3,3)$$ is

$$\displaystyle W=\int_0^3 { -\alpha x{ y }^{ 2 }.dy } \cos 180^{\circ} $$ and

$$ x= 3$$

$$\displaystyle W=\int_0^3 { \alpha \times 3 \times { y }^{ 2 }.dy } $$

$$\displaystyle W=2.5 \times 3 \times \int _{ 0 }^{ 3 }{ { y }^{ 2 }dy } $$

$$\displaystyle W=7.5\times\{ \dfrac { { 3 }^{ 3 } }{ 3 } -0\} $$

$$W=67.5 \ J $$
Question 6
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A particle of mass $$m$$ is suspended by a string of length $$l$$ from a fixed rigid support. A sufficient horizontal velocity $$\displaystyle v_{0}= \sqrt{3gl}$$ is imparted to it suddenly. Calculate the angle made by the string with the vertical when the acceleration of the particle is inclined to the string by $$\displaystyle 45^{\circ}.$$

A
$$\displaystyle \theta = \frac{\pi }{2}$$

B
$$\displaystyle \theta = \frac{\pi }{3}$$

C
$$\displaystyle \theta = \frac{\pi }{4}$$

D
$$\displaystyle \theta = \pi $$

Solution

Given : $$u = \sqrt{3gl}$$
Let the velocity of the bob be $$v$$ when it reaches the point C.
From figure, $$AB = l - lcos\theta = l(1-cos\theta)$$
Using work-energy theorem : $$W = \Delta K.E$$
$$\therefore$$ $$-mg l (1- cos\theta) = \dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2$$

OR $$-2 gl (1 - cos\theta) =v^2 - (3gl)$$ $$\implies v^2 = g l+ 2g lcos\theta$$

Using $$a_r =\dfrac{v^2}{l} = g + 2g cos\theta $$
Also tangential acceleration $$a_t =g sin\theta$$
As the acceleration makes an angle $$45^o$$ with the string, thus $$tan45^o = \dfrac{a_t}{a_r}$$
$$\implies$$ $$a_r = a_t$$
$$\therefore$$ $$g + 2g cos\theta = g sin\theta$$
$$\implies$$ $$sin\theta - 2cos\theta -1 = 0$$ where $$\theta =\dfrac{\pi}{2}$$ satisfies the equation.
Question 7
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The sphere at A is given a downward velocity $$\displaystyle v_{0}$$ of magnitude $$5 m/s$$ and swings in a vertical plane at the end of a rope of length $$l=2 m$$ attached to a support at $$O$$. Determine the angle $$\displaystyle \theta $$ at which the rope will break, knowing that it can withstand a maximum tension equal to twice the weight of the sphere.

A
$$\displaystyle \sin ^{-1}\left ( \frac{1}{4} \right )$$

B
$$\displaystyle \sin ^{-1}\left ( \frac{1}{3} \right )$$

C
$$\displaystyle \sin ^{-1}\left ( \frac{1}{2} \right )$$

D
$$\displaystyle \sin ^{-1}\left ( \frac{3}{4} \right )$$

Solution

Given : $$v_o =5 m/s$$ $$l =2$$ m $$T = 2mg$$
From figure, $$PM = l sin\theta$$
Circular motion equation at P : $$\dfrac{mv^2}{l} = T - mgsin\theta$$
OR $$mv^2 =l( 2mg - mgsin\theta) = 2 (2 m \times 10 - m(10) sin\theta)$$
$$\implies $$ $$mv^2 = m (40 - 20 sin\theta)$$

Work-energy theorem : $$W = \Delta K.E$$
$$\therefore$$ $$mg (PM) = \dfrac{1}{2}mv^2 - \dfrac{1}{2}mv_o^2$$

OR $$2mg (l sin\theta) =m(40 - 20sin\theta) - mv_o^2$$
OR $$2 m (10) (2 sin\theta) = m (40 - 20 sin\theta) - m(25)$$ $$\implies sin\theta = \dfrac{1}{4}$$
Question 8
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Directions For Questions

A small sphere B of mass $$m$$ is released from rest in the position shown and swings freely in a vertical plane, first about O and then about the peg A after the cord comes in contact with the peg. Determine the tension in the cord:

...view full instructions

just after it comes in contact with the peg.

A
$$\displaystyle \frac{mg}{2}$$

B
$$\displaystyle mg$$

C
$$\displaystyle \frac{3mg}{2}$$

D
$$\displaystyle \frac{5mg}{2}$$

Solution

Given : $$r = 0.8$$ m $$u = 0$$ m/s $$OA =0.4$$ m
$$\therefore$$ $$AM = r' =0.8 - 0.4 =0.4$$ m
After coming in contact with the peg, the sphere rotates in a circle of radius $$r' =0.4$$ m
Let the velocity of the sphere when it reaches the point v be $$v$$.
From figure, $$PC = r sin 30^o =0.4$$ m

Work-energy theorem : $$W = \Delta K.E$$
$$\therefore$$ $$mg (PC) =\dfrac{1}{2} mv^2 - \dfrac{1}{2}mu^2 $$

OR $$mg \times 0.4 = \dfrac{mv^2}{2} - 0$$
$$\implies mv^2 =0.8mg$$

Using: $$ ma_r = \dfrac{mv^2}{r'} = T' - mgsin30^o$$

$$\therefore$$ $$\dfrac{0.8mg}{0.4} = T' - \dfrac{mg}{2}$$
$$\implies T' = \dfrac{5mg}{2}$$
Question 9
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Directions For Questions

A small sphere B of mass $$m$$ is released from rest in the position shown and swings freely in a vertical plane, first about O and then about the peg A after the cord comes in contact with the peg. Determine the tension in the cord:

...view full instructions

just before the sphere comes in contact with the peg.

A
$$\displaystyle \frac{mg}{2}$$

B
$$\displaystyle {mg}$$

C
$$\displaystyle \frac{3mg}{2}$$

D
$$\displaystyle \frac{5mg}{2}$$

Solution

Given : $$r = 0.8$$ m $$u = 0$$ m/s
Before coming in contact with the peg, the sphere rotates in a circle of radius $$r =0.8$$ m
Let the velocity of the sphere when it reaches the point C be $$v$$.
From figure, $$PC = r sin 30^o = \dfrac{r}{2}$$ m

Work-energy theorem : $$W = \Delta K.E$$
$$\therefore$$ $$mg (PC) =\dfrac{1}{2} mv^2 - \dfrac{1}{2}mu^2 $$

OR $$mg \times \dfrac{r}{2} = \dfrac{mv^2}{2} - 0$$

$$\implies \dfrac{mv^2}{r} =mg$$

Using: $$ ma_r = \dfrac{mv^2}{r} = T - mgsin30^o$$
$$\therefore$$ $$mg = T - \dfrac{mg}{2}$$
$$\implies T = \dfrac{3mg}{2}$$
Question 10
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Directions For Questions

Bob B of the pendulum AB is given an initial velocity $$\displaystyle \sqrt{3Lg}$$ in horizontal direction. Find the maximum height of the bob from the starting point:

...view full instructions

if AB is a massless rod,

A
$$\displaystyle \frac{L}{2}$$

B
$$\displaystyle \frac{3L}{2}$$

C
$$\displaystyle \frac{5L}{2}$$

D
$$\displaystyle \frac{7L}{2}$$

Solution

Initial velocity of the rod $$u = \sqrt{3Lg}$$
Let the maximum height attained to be $$H$$. At the starting point, its potential energy is zero while at the highest point, its kinetic is zero.
Applying conservation of energy : $$P.E_i + K.E_ i =P.E_f + K.E_f$$
$$\therefore$$ $$0 + \dfrac{1}{2}mu^2 = mgH + 0$$

$$\implies$$ $$H = \dfrac{u^2}{2g} = \dfrac{3Lg}{2g} = \dfrac{3L}{2}$$