Work Energy and Power Test

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Work Energy and Power Test - 30

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Question 1
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In a vertical circle of radius $$(r)$$, at what point in its path a particle may have tension equal to zero:

A
highest point

B
lowest point

C
at any point

D
at a point horizontal from the centre of radius

Solution

Let the tension in the string at the highest point be $$T$$
Minimum speed required by the particle at the highest point to complete the vertical circular motion is $$\sqrt{gr}$$
$$\therefore$$ $$\dfrac{mv^2}{r} =T + mg$$

OR $$\dfrac{m (gr)}{r} = T+mg$$ $$\implies T =0$$
Thus tension can be zero at the highest point.
Question 2
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A stone attached to one end of a string is whirled in a vertical circle. The tension in the string is maximum when

A
the string is horizontal

B
the string is vertical with the stone at highest position

C
the string is vertical with the stone at the lowest position

D
the string makes an angle of $$45^{\circ}$$ with the vertical

Solution

$$Answer:-$$ C
When the stone is at the bottom position forces acting on stone are:
Upward force tension (T), Downward forces are weight (mg) and also $$\dfrac{mv^2}{R}$$ centrifugal force
So balancing forces we get :
$$T=mg+\dfrac{mv^2}{R}$$
And minimum tension is obtained at highest point which will be:
$$T=mg-$$ $$\dfrac{mv^2}{R}$$
Question 3
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A position dependent force $$F=7-2x+3x^{2}$$N acts on a small object of mass 2kg to displace it from $$x=0$$ to $$x=5m$$. The work done in joules is-

A
70

B
270

C
35

D
135

Solution

$$W=\int F\mathrm{d} x$$
$$\displaystyle \int_{0}^{5}(7-2x+3x^{2})\mathrm{d} x=\left ( 7x-\frac{2x^{2}}{2}+\frac{3x^{3}}{3} \right )_{0}^{5}=135J$$
Question 4
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A particle is moving in a vertical circle, the tension in the string when passing through two position at angle $$30^{\circ}$$ and $$60^{\circ}$$ from vertical from lowest position are $$T_1$$ and $$T_2$$ respectively, then

A
$$T_1 = T_2$$

B
$$T_1 > T_2$$

C
$$T_1 < T_2$$

D
$$T_1\ge T_2$$

Solution

In equilibrium, $$T=mv^2/r+mg\cos\theta$$
Since, $$\theta$$ increases $$\cos\theta$$ and $$v$$ both will decrease.
Hence, for $$\theta=60^o$$, the tension will be less.
Thus, $$T_1>T_2$$.
Question 5
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A pendulum bob has a speed $$3\;m/s$$ while passing through its lowest position, length of the pendulum is $$0.5\;m$$ then its speed when it makes an angle of $$60^{\circ}$$ with the vertical is

A
$$\;2\;m/s$$

B
$$\;1\;m/s$$

C
$$\;4\;m/s$$

D
$$\;3\;m/s$$

Solution

$$\textbf{Step 1: Apply energy conservation}$$
In $$\Delta ABD$$,

$$AB = AD \cos \theta$$

$$\Rightarrow AB = l \cos \dfrac{\pi}{3} = \dfrac{l}{2}$$

$$\Rightarrow h = l - \dfrac{l}{2} = \dfrac{l}{2}$$

As there are no dissipative force energy will be conserved.
$$\Rightarrow (E)_i = E_f$$

$$\Rightarrow K_1 + U_1 = K_2 + U_2$$
Taking point $$C$$ is zero potential level

$$\Rightarrow \dfrac{1}{2} mu^2 = \dfrac{1}{2} mv^2 + (mgh)$$

$$\textbf{Step 2: Calculations}$$

$$\Rightarrow v = \sqrt{u^2 - 2gh}$$

Putting given values
$$v = \sqrt{(3)^2(m/s)^2 - 2 \times 9.8(m/s^2) \times \dfrac{0.5}{2}(m)} = 2 m/s$$

Hence $$A$$ is the correct option.
Question 6
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A particle slides from rest from the topmost point of a vertical circle of radius ($$r$$) along a smooth chord making an angle $$(\theta)$$ with the vertical the time of decent is

A
Least for $$\theta=0^{\circ}$$

B
Maximum for $$\theta=0^{\circ}$$

C
Least for $$\theta=45^{\circ}$$

D
Independent of $$\theta$$

Solution

Force on particle along the cord = $$mg\cos { \theta }$$
Distance travelled by the particle =$$\quad \dfrac { d }{ \cos { \theta } }$$, where $$d$$ is the diameter or the vertical circle.

$$s = \dfrac { a{ t }^{ 2 } }{ 2 }$$

$$\Rightarrow\ t =\sqrt { \dfrac { 2s }{ a } }$$.

Hence,$$\ t$$ is independent of $$\theta$$.
Question 7
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A particle of mass $$m_{1}$$ moving with a velocity of $$5m/s$$ collides head on with a stationary particle of mass $$m_{2}$$. After collision both the particle move with a common velocity of $$4m/s$$, then the value $$m_
{1}/m_{2}$$ is:

A
4:1

B
2:1

C
1:8

D
1:2

Solution

Conservation of Momentum principle,
Initial momentum is $$M_i=5m_1$$
Final momentum is $$M_f=4(m_1+m_2)$$
By the above stated principle $$M_i=M_f=5m_1=4(m_1+m_2)$$
$$\implies m_1=4m_2$$
$$\therefore m_1:m_2=4:1$$
Question 8
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Under the action of a force, a 2kg mass moves such that its position x as a function of time t is given by $$x=t^3/3$$ where x is in meters and t in second. The work done by the force in first two seconds is:

A
1600 joule

B
160 joule

C
16 joule

D
1.6 joule

Solution

$$x=\dfrac{t^3}{3}$$
$$ a=\dfrac{d^2x}{dt^2}=2t$$
Thus force acting on the particle=$$ma=2mt$$
Work done$$=\int Fdx$$
$$=\int_0^2 (2mt)t^2dt$$
$$=2m\int_0^2 t^3dt$$
$$=16J$$
Question 9
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Force shown acts for $$2$$ seconds. Find out work done by force $$F$$ on $$10$$ kg in $$3$$ seconds.

A
30 J

B
20 J

C
50 J

D
60 J

Solution

Work done,
$$W=Fd$$
Displacement $$d$$ is given by,
$$d=\dfrac{1}{2}a{t}^{2}$$
$$F=ma$$
$$10=10a$$, $$a=1m/s^2$$
$$d=\dfrac{1}{2}(1){(2)}^{2}=2m$$
$$W=10\times2=20J$$
Question 10
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A time dependent force F = 10 t is applied on 10 kg block. Find out the work done by F in 2 seconds.

A
40 J

B
20 J

C
120 J

D
60 J

Solution

$$10\, \times\, \displaystyle \dfrac{dv}{dt}\, =\, 10\, t$$

$$\int_0^v \ dv =\int_{0}^{t} tdt$$

$$\Rightarrow\, v\, =\, \displaystyle \dfrac{t^2}{2}$$ ..........(1)

$$dW\, =\, \vec{F}\, .\, \vec{ds}$$
$$dW\, =\, 10 t. dx$$
$$dW\, =\, 10 t. v. dt$$ .......... (2) $$\because\, dx\, =\, vdt$$

from (1) & (2)

$$dW\, =\, 10\, t\, .\, \left ( \displaystyle \dfrac{t^2}{2} \right )dt\, \quad\, dW\, =\, 5t^3dt$$

$$W\, =\, \displaystyle \dfrac{5}{4}\, [t^4]_0^2\, =\, 20\, J$$