Work Energy and Power Test

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Work Energy and Power Test - 53

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Question 1
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Mass of $$B$$ is four times that of $$A . B$$ moves with a velocity half that of $$A$$. Then, $$B$$ has

A
kinetic energy equal to that of $$A$$

B
half the kinetic energy of $$A$$

C
twice the kinetic energy of $$A$$

D
kinetic energy one-fourth of $$A$$

Solution

According to question given,
$$M_B = 4 \times M_A $$ and $$ V_B = \dfrac{V_A}{2} $$

$$K.E_A = \dfrac{1}{2}M_AV_A^2 $$ .........(1)

$$K.E_B = \dfrac{1}{2}M_BV_B^2 $$ = $$\dfrac{1}{2}\ (4 \times M_A) \left(\dfrac{V_A}{2}\right)^2=\dfrac{1}{2}M_AV_A^2 $$........(2)

From above equation
$$K.E_A=K.E_B$$
Question 2
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A body of mass $$15\ kg$$ moving with a velocity of $$10\ ms^{-1}$$ is bought to rest. The work done by the brake is

A
$$-250\ J$$

B
$$-500\ J$$

C
$$-750\ J$$

D
$$-1000\ J$$

Solution

$$W = change\ in\ kinetic\ energy =\dfrac{1}{2}m(v^2-u^2) $$

$$W = \dfrac{1}{2} \times 15 \times (0^2-10^2)$$

$$W = - 750J$$

Hence, option $$C$$ is correct answer.
Question 3
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The work done in placing a charge of 8$$\times $$$${ 10 }^{ -18 }$$ coulomb on a condenser of capacity 100 micro-farad is

A
16$$\times { 10 }^{ -32 }$$ joule

B
3.2$$\times { 10 }^{ -26 }$$ joule

C
4$$\times { 10 }^{ -10 }$$ joule

D
32$$\times { 10 }^{ -32 }$$ joule

Solution
Question 4
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A bod of mass M is suspended by a massless string of length L.The horizontal velocity V at position A is just sufficient to make real the point B.The angle $$\theta$$ at which the speed of the bob is half of that it A satisfies:

A
$$\theta=\dfrac{\pi}{4}$$

B
$$\dfrac{\pi}{4} < \theta<\frac{\pi}{2}$$

C
$$\dfrac{\pi}{2} < \theta<\dfrac{3\pi}{4}$$

D
$$\dfrac{3\pi}{4} < \theta<\pi$$

Solution

Use conservation of energy,
$$\cfrac { 1 }{ 2 } m{ v }_{ o }^{ 2 }=\cfrac { 1 }{ 2 } m{ v }^{ 2 }+mgl \left( 1-\cos { \theta } \right) -------(i)$$
where $$v_{o}=$$horizontal velocity
$$mg(2l)=\cfrac { 1 }{ 2 } m{ v }_{ o }^{ 2 }-\cfrac { 1 }{ 2 } m{ v }_{ top }^{ 2 }-------(ii)$$
Since $$v_{o}$$ is just sufficient
$$\cfrac { m{ v }_{ top }^{ 2 } }{ l } =T+mg\\ T=0,{ v }_{ top }=\sqrt { gl } $$
Then equation $$(ii)$$,
$$\Longrightarrow { v }_{ o }=\sqrt { 5gl } $$
According to the question, $$v=\cfrac { { v }_{ o } }{ 2 } $$
From equation $$(i),$$
$$\cfrac { 1 }{ 2 } m5gl=\cfrac { 1 }{ 2 } m\left( \cfrac { 5gl }{ 4 } \right) +mgl\left( 1-\cos { \theta } \right) \\ \Longrightarrow \cfrac { 20mgl-5mgl }{ 8 } =mgl\left( 1-\cos { \theta } \right) \\ \Longrightarrow \cos { \theta } =\cfrac { 7 }{ 8 } $$
Hence, $$\cfrac{3 \pi}{4} < \theta < \pi$$.
Question 5
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A body of mass 1 kg begins to move under the action of a time dependent force $$ \overrightarrow { F } =(2t\hat { i } +3t^{ 2 }\hat { j } ) $$ N, where $$ \hat { i } $$ and $$ \hat { j } $$ are unit vectors along x and y axes. What power will be developed by the force at the time t?

A
$$ (2t^2+3t^2)W $$

B
$$ (2t^4+4t^4)W $$

C
$$ (2t^3+3t^4)W $$

D
$$ (2t^3+3t^5)W $$

Solution

As given $$\overrightarrow{F}=(2t\hat{i}+3t^2\hat{j})N$$
Where $$\hat{i}$$ and $$\hat{j}$$ are unit vectors along x and y axis
$$F=ma$$
$$\Rightarrow a=\cfrac{F}{m}=\cfrac{(2t\hat{i}+3t^2\hat{j})}{2}=[m=1kg]\\ \Rightarrow a=(2t\hat{i}+3t^2\hat{j})m/s^2$$
Acceleration $$a=\cfrac{dv}{dt}$$
$$\Rightarrow dv=a.dt\rightarrow(i)\\ \int{dv}=\int{a.dt}=\int{(2t\hat{i}+3t^2\hat{j})}dt\\v=t^2\hat{i}+t^3\hat{j}\\P=F.v=(2t\hat{i}+3t^2\hat{j})(t^2\hat{i}+t^3\hat{j})\\ \quad=2t.t^+3t^2t^3\\P=(2t^3+3t^5)W$$
Question 6
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A particle is moving in a vertical circle. The tension in the string when passing through two positions at angles $${30}^{o}$$ and $${60}^{o}$$ from vertical (lowest position) are $${T}_{1}$$ and $${T}_{2}$$ respectively, then

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

B
$${T}_{2}> {T}_{1}$$

C
$${T}_{1}> {T}_{2}$$

D
tension in the string always remains the same

Solution

From the Free Body Diagram we get,
$$T= \dfrac{mv^2}{r}+ mg cos \theta$$
At $$ \theta= 30 ^0 \quad T = T_1= \dfrac{mv^2}{r}+ \dfrac{\sqrt{3}mg}{2} $$
At $$\theta= 60 ^0 \quad T= T_2= \dfrac{mv^2}{r}+ \dfrac{mg}{2}$$
From Above we get
$$T_1 > T_2$$
Question 7
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A body of mass $$2\ kg$$ moving under a force has relation between displacement $$x$$ and time $$t$$ as $$x=\dfrac{t^{3}}{3}$$ where $$x$$ is in metre and $$t$$ is in sec. The work done by the body in first two second will be

A
$$1.6\ joule$$

B
$$16\ joule$$

C
$$160\ joule$$

D
$$1600\ joule$$

Solution

work done = change in kinetic energy

$$x=\dfrac{t^3}{3}$$
velocity $$v=\dfrac{dx}{dt}=\dfrac{d\left ( \dfrac{t^3}{3} \right )}{dt}=t^2$$

given, $$t=2s$$

Velocity $$=t^2=2^2=4$$

$$W=\dfrac{1}{2}mv^2$$

$$=\dfrac{1}{2} \times 2 \times 4^2$$

$$=16J$$
Question 8
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Work done by a force $$\vec{F}=(\hat{i}+2\hat{j}+3\hat{k})\ N$$ action on a particle in displacing it from the point $$\vec{r}_{1}=\hat{i}+\hat{j}+\hat{k}$$ to the point $$\vec{r}_{1}=\hat{i}-\hat{j}-2\hat{k}$$

A
$$-1\ J$$

B
$$2\ J$$

C
$$-13\ J$$

D
$$zero$$

Solution

work = force * displacement,

$$W=F\times s$$

$$s=(i-j-2k)-(i+j+k)=-2j-3k$$

$$F=i+2j+3k$$

$$W=(i+2j+3k)\times (-2j-3k)$$

$$\therefore W=-13J$$
Question 9
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By applying a force $$F=(3xy-5z) j +4zk$$ a particle is moved along the path $$y=x^2$$ from point $$(0,0,0)$$ to the point $$(2,4,0)$$. The work done by the F on the particle is (all values are in SI units)

A
$$\dfrac { 280 }{ 5 } J$$

B
$$\dfrac { 140 }{ 5 } J$$

C
$$\dfrac { 232 }{ 5 } J$$

D
$$\dfrac { 192 }{ 5 } J$$

Solution

$$ \bar{F} = (3xy-5z)\hat{j}+4z\hat{k} $$
$$ x^{2} = y $$
$$ x = y ^{1/2} $$
Substitute in above equation
$$ \bar{F} = (3y^{3/2}-5z)\hat{j}-4z\hat{k} $$
dw = F.dx
In $$ (\hat{j}) $$ direction
$$ W = _{0}^{4}\int 3y^{3/2}dy $$
$$ \rightarrow \frac{3y^{5/2}}{5/2}|_{0}^{4} $$
$$ W = \frac{192}{5} $$ joule
(in 'z' work done is always zero)
Question 10
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Two masses $$10 \ gm$$ and $$40 \ gm$$ are moving with kinetic energies in the ratio $$9:25$$. The ratio of their linear momentum is:

A
$$5:6$$

B
$$3:10$$

C
$$6:5$$

D
$$10:3$$

Solution