Work Energy and Power Test

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Work Energy and Power Test - 22

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Question 1
1 / -0

A stone of mass $$1 kg$$ tied to a light inextensible string of length $$L=\dfrac{10}{3}m$$ is whirling in a circular path of radius $$L$$ in a vertical plane. If the ratio of the maximum tension in the string to the minimum is $$4$$ and if $$g$$ is taken to be $$10 \ m/s^2$$, then speed of stone at the highest point of the circle is

A
$$20 m/sec$$

B
$$10\sqrt{3} \ m/sec$$

C
$$5\sqrt{2} \ m/sec$$

D
$$10 m/sec$$

Solution

Let '$$v$$' be the velocity at the top point
Velocity at the bottom point is $$= v^2 + 2gd $$ $$ = v^2 + 4gr$$
Maximum tension in string appears when the stone is at the bottom of the circle.

$$ T_{max} = \dfrac{m(v^2 + 4gr)}{r} + mg $$

Minimum tension appears in a string when the stone is at the top of the circle.

$$ T_{min} = \dfrac{mv^2}{r} - mg$$

$$\dfrac{T_{max}}{T_{min}} = \dfrac{\dfrac{v^2 + 4gr}{r} + g}{\dfrac{v^2}{r} - g}$$ $$ = 4$$

On solving the equation we get, $$v $$$$= 10\ m/s $$
Question 2
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A car is accelerated on a leveled road and attains a velocity 4 times of its initial velocity. In this process the potential energy of the car

A
does not change

B
becomes twice to that of initial

C
becomes 4 times that of initial

D
becomes 16 times that of initial

Solution

Potential energy is the energy possessed by a body by the virtue of its height from the ground and in the given case, the car only increases its velocity 4 times but its height from the ground remains constant. Thus, the potential energy does not change.
Question 3
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An 800 g ball is pulled up a slope as shown in the diagram. Calculate the potential energy it gains.

A
1.96 J

B
1.568 J

C
7.84 J

D
1.225 J

Solution
Question 4
1 / -0

A pendulum hangs from the ceiling of a jeep moving with a speed, $$v$$, along a circle of radius, $$R$$. Find the angle with the vertical made by the pendulum.

A
$$0$$

B
$$\tan ^{-1}\displaystyle \frac{v^{2}}{Rg}$$

C
$$\tan ^{-1}\displaystyle \frac{Rg}{v^{2}}$$

D
None of the above

Solution

Considering accelerations,
$$a_{r}=\displaystyle \dfrac{v^{2}}{R}$$
$$\therefore tan\theta =\displaystyle \dfrac{v^2/R}{g}$$
$$=\displaystyle \dfrac{v^{2}}{Rg}$$
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 $$\dfrac {v_0}{\sqrt 2}$$. Then the speed of the particle of mass $$2m_0$$ after this collision is

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

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

C
$$\sqrt 2v_0$$

D
$$\dfrac {v_0}{\sqrt 2}$$

Solution

From conservation of momentum, $$\vec{p_i}=\vec{p_f}$$;
Since the particle of mass $$m_0$$ moves at angle $$45^0$$, we have the unit vector $$(\dfrac{1}{\sqrt2}\hat{i}+\dfrac{1}{\sqrt2}\hat{j})$$ to show the direction.
$$\therefore m_0v_0\hat{i}=m_0\dfrac{v_0}{\sqrt 2}(\dfrac{1}{\sqrt2}\hat{i}+\dfrac{1}{\sqrt2}\hat{j}) +2m_0\vec{v}$$

$$\therefore \vec{v}=\dfrac{v_0}{4}\hat{i}-\dfrac{v_0}{4}\hat{j}$$

$$\therefore\left |\vec{v} \right |=\dfrac{v_0}{2\sqrt2}$$
Question 6
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what is the work done by a force $$4N$$ in moving the body from d $$=$$ 1m to 4m?

A
6 J

B
16 J

C
12 J

D
8 J

Solution

$$W=F.S \rightarrow 4(4-1)J = 12J$$
Question 7
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A spring of natural length $$l$$ is compressed vertically downward against the floor so that its compressed length becomes $$\displaystyle\frac{l}{2}$$. On releasing, the spring attains it's natural length. If $$k$$ is the stiffness constant of the spring, then the work done by the spring on the floor is

A
$$zero$$

B
$$\displaystyle\frac{1}{2}kl^{2}$$

C
$$\displaystyle\frac{1}{2}k(\displaystyle\frac{l}{2})^{2}$$

D
$$kl^{2}$$

Solution

Only the center of mass of the spring changes, but the floor has no displacement.
$$W=F.x$$
$$W=F\times0$$
$$W=0$$
Hence the work done by the spring on the floor is zero.
Option A.
Question 8
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A heavy ball moving with speed $$v$$ collides with a tiny ball. The collision is elastic, then immediately after the impact, the second ball will move with a speed approximately equal to-

A
$$v$$

B
$$2v$$

C
$$\dfrac {v}{2}$$

D
$$\dfrac {v}{3}$$

Solution

Using the conservation of momentum and KE, the velocities of the balls after collision is derived.
$$m_1=$$ mass of   the   1st    ball
$$m_2=$$ mass   of   the   2nd   ball
$$v_1=$$ velocity   of   the  1st   ball before collision$$=v$$;
$$v_2=$$ velocity   of   the   2nd   ball before collision$$=0$$

$${v_2}^{'}= \left(\dfrac{2m_1}{m_1+m_2}\right) v_1 + \left(\dfrac{m_1-m_2}{m_1+m_2}\right)v_2$$;

$$m_1 > > m_2$$;
Substituting $$v_1=v$$   and $$v_2=0$$,we get   $${v_2}'=2v $$
Question 9
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A particle moves along the x-axis from $$x=0$$ to $$x=5$$ m under the influence of a force given by $$F=7-2x+3x^2$$. The work done in the process is

A
360 J

B
85 J

C
185 J

D
135 J

Solution

Work done is given by,
$$\displaystyle W=\int_0^5 Fdx$$

$$\displaystyle =\int_0^5(7-2x+3x^2)dx$$

$$=7x-x^2+x^3$$

$$=135 \ J $$
Question 10
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A simple pendulum is oscillating without damping. When the displacement of the bob is less than maximum, its acceleration vector $$\overrightarrow{a}$$ is correctly shown in

A

B

C

D

Solution

The bob has both radial as well as tangential acceleration when it is at a displacement less than its maximum displacement.
From figure $$a_t =g sin\theta$$and $$a_r = \dfrac{T}{m} - g cos\theta$$
Thus resultant acceleration $$\vec{a} = \vec{a_r} + \vec{a_t}$$ points in the direction as shown in the figure.