Work Energy and Power Test

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Work Energy and Power Test - 47

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Question 1
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A body of mass travels in a straight line with a velocity $$v=kx^{3/2}$$ where $$k$$ is a constant. The work done in displacing the body from $$x=0$$ to $$x$$ is proportional to:

A
$$x^{1/2}$$

B
$$x^{2}$$

C
$$x^{3}$$

D
$$x^{5/2}$$

Solution

Given,

Velocity, $$\dfrac{dx}{dt}=v=k{{x}^{3/2}}$$

Acceleration,

$$ a=\dfrac{dv}{dt}=\dfrac{d\left( k{{x}^{3/2}} \right)}{dt}=\dfrac{1}{2}k{{x}^{1/2}}\dfrac{dx}{dt}=\dfrac{1}{2}k{{x}^{1/2}}v $$

$$ a=\dfrac{1}{2}k{{x}^{1/2}}\left( k{{x}^{3/2}} \right)=\dfrac{1}{2}{{k}^{2}}{{x}^{2}} $$

$$ work\,done\,=\,force\times displacemet\,=Mass\times acceleration\times displacement $$

$$ \Rightarrow dW=m.a\,.dx=m.\dfrac{1}{2}{{k}^{2}}{{x}^{2}}dx $$

On integrating

$$W=m\dfrac{1}{2}{{k}^{2}}\left( \dfrac{{{x}^{3}}}{3} \right)$$

Hence, Work done is proportional to $${{x}^{3}}$$
Question 2
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A force $$\vec { F }=(3t\hat { i } +5\hat { j })$$N acts on a body due to which its position varies as $$\vec { S }=(2{t^2}\hat { i } -5\hat { j })$$. Find the work done by this force in initial $$2s$$.

A
$$23\ J$$

B
$$32\ J$$

C
$$Zero$$

D
$$48\,J$$

Solution

Given,

Force, $$F=\left( 3t\hat{i}+5\hat{j} \right)N$$

Displacement, $$s=\left( 2{{t}^{2}}\hat{i}-5\hat{j} \right)\,m$$

$$ds=\left( 4t\hat{i}+0\hat{J} \right)\,dt$$

$$ dw=F.ds $$

$$ \int_{0}^{w}{dw}=\int_{0}^{2}{\left( 3t\hat{i}+5\hat{j} \right)}\left( 4t\hat{i}+0\hat{j} \right)\,dt $$

$$ W=\left. 12{{t}^{2}} \right|_{0}^{2}=48\,J $$

Net work done is $$48\,J$$
Question 3
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A particle moves from origin to position $$\vec {r}_{1}=3\hat {i}+2\hat {j}-6\hat {k}$$ under the action of force $$4\hat {i}+\hat {j}+3\hat {k}N$$. the work done will be

A
$$10\ J$$

B
$$5\ J$$

C
$$2\ J$$

D
$$-4\ J$$

Solution
Question 4
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A body of mass $$2kg$$ makes an elastic collision with another body at rest and comes to rest .The mass of the second body which collides with the first body is

A
$$2 kg$$

B
$$1.2 kg$$

C
$$3 kg$$

D
$$1 kg$$

Solution
Question 5
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Under influence of a force $$\overrightarrow { F } \left( x \right) =\left( 3{ x }^{ 2 }-2x+5 \right) \hat { i } N$$ there is displacement of a particle from $$x=0$$ and $$x=5m$$ on $$x-$$axis. So work done is $$J$$.

A
$$100$$

B
$$150$$

C
$$125$$

D
$$120$$

Solution
Question 6
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A particle moves under the effect of a force $$F=c\ x$$ from $$x=0$$ to $$x=x_{1}$$. The work done in the process is

A
$$\dfrac{c{ { x }_{ 1 } }^{ 2 }}{2}$$

B
$$c{ { x }_{ 1 } }^{ 2 }$$

C
$$c{ { x }_{ 1 } }^{ 3 }$$

D
$$Zero$$

Solution
Question 7
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An inclined track ends in a circular loop of diameter $$D$$. From what height on the track a particles should be released so that it completes that loop in the vertical plane?

A
$$\dfrac {5D}{2}$$

B
$$\dfrac {2D}{5}$$

C
$$\dfrac {5D}{4}$$

D
$$\dfrac {4D}{5}$$

Solution
Question 8
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A ball is dropped from certain height, after striking the gourn it rebounds till $$\frac{2}{5}th$$ of the initial height. The ration of its speed just before and just after striking the ground is [assume no loss of energy due to air friction]

A
$$\sqrt{\dfrac{2}{5}}$$

B
$$\dfrac{2}{5}$$

C
$$\sqrt{\dfrac{5}{2}}$$

D
$$\dfrac{5}{2}$$

Solution
Question 9
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A force acts on a $$30 gm$$ particle in such a way that the position of the particle as a function of time is given by $$x=3t-4{t}^{2}+{t}^{3}$$, where $$x$$ is in metres and $$t$$ is in seconds. The work done on the particle during the first $$4$$ second is

A
$$3.84J$$

B
$$1.68J$$

C
$$5.28J$$

D
$$5.41J$$

Solution

Given,
$$m=30g=\dfrac{30}{1000}=0.03kg$$
$$x=3t-4t^2+t^3$$
$$dx=(3-8t+3t^2)dt$$
Velocity, $$v=\dfrac{dx}{dt}$$
$$v=3-8t+3t^2$$
Acceleration, $$a=\dfrac{dv}{dt}$$
$$a=-8+6t$$
Work done on the particle during the first $$4$$ second
$$W=\int F.dx=m\int adx$$
$$W=m\int_0^4 (-8+6t)(3-8t+3t^2)dt$$
$$W=0.03\int _0^4 (18t^3-72t^2+82t-24)dt$$
$$W=0.03[\dfrac{18}{4}t^4-\dfrac{72}{3}t^3+\dfrac{82}{2}t^2-24t]_0^4$$
$$W=0.03\times 176$$
$$W=5.28J$$
The correct option is C.
Question 10
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A stone, tied at the end of a string $$80$$ cm long, is whirled in a horizontal circle with a constant speed. If the stone makes $$14$$ revolutions in $$25$$ sec, what is the magnitude of acceleration of the stone.

A
$$680$$ cm$$/s^2$$

B
$$720$$ cm$$/s^2$$

C
$$860$$ cm$$/s^2$$

D
$$990$$ cm$$/s^2$$

Solution

Given length of string = 80 cm and angular velocity $$\omega $$ = 2$$\pi $$f=$$\dfrac { 88 }{ 25 } $$
And we know centripetal acceleration =$$r{ \omega }^{ 2 }$$
a=$$80\times { (\frac { 88 }{ 25 } ) }^{ 2 }$$
=991.232 cm/s
=990 cm/s