Work Energy and Power Test

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Work Energy and Power Test - 64

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Question 1
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Two blocks A and B of masses m and 2m respectively placed on a smooth floor are connected by a spring. A third body C of mass m moves with velocity $$v_0$$ along the line joining A and B and collides elastically with A. At a certain instant of time after collision it is found that the instantaneous velocities of A and B are same then :

A
the common velocity of A and B at time t$$_0$$ is v/3.

B
the spring constant is k $$= \displaystyle \frac{3mv_0^2}{2x_0^2}$$.

C
the spring constant is k $$= \displaystyle \frac{2mv_0^2}{3x_0^2}$$.

D
none of these

Solution

Since the masses of A and C are same and the collision is elastic, their velocities are exchanged after the collision. So, now the system is (A+B) and initially B is at rest wheras, A has velocity $${ v }_{ 0 }$$. After some time they have same velocities, so conserving momentum
$${ mv }_{ 0 }=(m+2m)v$$

$$v =\dfrac { { v }_{ 0 } }{ 3 } $$
Question 2
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A force $$F = - K (y \widehat i + x \widehat j)$$ (where $$K$$ is positive constant) acts on a particle moving in the $$x-y$$ plane. Starting from the origin, the particle is taken along the $$x$$-axis to the point $$(a, 0)$$ and then parallel to $$y$$-axis to the point $$(0, a)$$. The total work done by the force $$F$$ on the particle is:

A
$$- 2 Ka^2$$

B
$$2 Ka^2$$

C
$$- Ka^2$$

D
$$Ka^2$$

Solution

Given
$$\vec F = -K(y \vec i+x \vec j)$$
Consider the small work done by the force $$F$$ in displacing the particle through area $$dS$$ as

$$dW = \vec F.\vec {dS}$$

where, $$\vec {dS} = dx \hat{i} + dy \hat j$$

$$\therefore dW = ( -K(y \hat i+x \hat j)).( dx \hat i + dy \hat j)$$

$$\therefore dW = -K(y dx + x dy) = - Kd(xy)$$ .....................(1)

Since, first the particle is taken along the position x-axis to the point (a, 0) and then parallel to y-axis to the point (0, a) hence, integrating for whole area we get

$$\displaystyle W = \int_{x=0}^{x=a} \int_{y=0}^{y=a} dW$$

$$\displaystyle W = \int_{x=0}^{x=a} \int_{y=0}^{y=a} (- Kd(xy))$$.............from(1)

$$\displaystyle W = -K(\left | x \right |_{0}^{a})(\left | y \right |_{0}^{a})$$

$$W = -Ka^2$$
Question 3
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Directions For Questions

Refer figure, which shows a ball of weight $$W$$ hanging at the end of a light and inextensible string of length $$L$$. A force $$F$$, which is always kept horizontal, is applied to push the ball sideways, and its value is very slowly raised from $$0$$ to $$F$$ corresponding to rise h of the ball (angle $$\theta$$). Such a process is called quasi static process. The force is increased so slowly that the ball may be treated to be in equilibrium under the forces $$T, W$$ and $$F$$ all the time, where $$T$$ is the tension in the string. Now answer the following questions.

...view full instructions

The work done by the varying force in changing the angular displacement from 0 to $$\theta $$ is

A
$$Wh$$

B
$$FL\,\sin \theta$$

C
$$Fh$$

D
$$\dfrac{1}{2}FL\,\sin \theta $$

Solution

Because the process is quasi static, work is equal to increase of potential energy.
$$Work =mgh= Wh$$.
Question 4
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Directions For Questions

Your physics teacher who gives you tuitions asks you to do experiments on the springs which do not follow Hooke's law faithfully. He gives you a spring and asks to fix one end. Connect a mass $$M$$ on the other end (The system is on a smooth table). (i) The spring is stretched by $$l$$ and released. (ii) Then he asks you to bring another identical block and provide velocity $$v$$ so that it pushes the spring by $$x$$. The spring follows the law $$F = kx - rx^2$$. The teacher asks you to find:

...view full instructions

The work done by the spring in case (i):

A
$$\displaystyle \frac{kl^2}{2} - \frac{rl^3}{3}$$

B
$$\displaystyle \frac{kl^2}{2} + \frac{rl^2}{3}$$

C
$$\displaystyle \frac{kl^3}{6} - \frac{rl^2}{8}$$

D
$$\displaystyle \frac{kl^3}{8} - \frac{rl^2}{6}$$

Solution

Given, $$F=$$$$kx-rx^{2}$$
workdone by spring in case(i) $$=$$$$\int_{0}^{l}Fdx$$
$$ =-$$[$$\dfrac{kl^{2}}{2} -\dfrac{rl^{3}}{3} $$] {since, $$F$$ and displacement are in opposite direction }
Question 5
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A force $$F = - K (y \widehat i + x \widehat j)$$, (where $$K$$ is a +ve constant) acts on a particle moving in xy plane starting from origin, the particle is taken along the positive x-axis to the point ($$a, 0$$) and then parallel to y axis to the point ($$a, a$$). The total work done by force $$F$$ on the particle is

A
$$- 2 Ka^2$$

B
$$2 Ka^2$$

C
$$- Ka^2$$

D
$$ Ka^2$$

Solution

Work was done $$\int Fdt$$

When particle moves from $$0$$ to $$a$$ along x-axis

Then $$W_1 = \displaystyle \int_0^a - (Ky \widehat i) dx = 0$$ $$\because\,\, y=0 $$

Now, when the particle moves parallel to the y-axis

$$W_2 = \displaystyle \int_0^a - (Ka \widehat j) dy $$ $$\because\,\, x=a $$

$$\therefore W_2= - Ka \displaystyle \int_0^a dy = - Ka^2$$

$$\therefore W = W_1 + W_2 = - Ka^2$$
Question 6
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Directions For Questions

Refer figure, which shows a ball of weight $$W$$ hanging at the end of a light and inextensible string of length $$L$$. A force $$F$$, which is always kept horizontal, is applied to push the ball sideways, and its value is very slowly raised from $$0$$ to $$F$$ corresponding to rise h of the ball (angle $$\theta$$). Such a process is called quasi static process. The force is increased so slowly that the ball may be treated to be in equilibrium under the forces $$T, W$$ and $$F$$ all the time, where $$T$$ is the tension in the string. Now answer the following questions.

...view full instructions

The work done by the tension T in the above process is

A
$$Zero$$

B
$$T(L-L\cos \theta )$$

C
$$-TL$$

D
$$-TL\,\sin \theta $$

Solution

The instantaneous displacement is directed normal to $$T$$ all the time.
Therefore $$W_T=0$$.
Question 7
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Two springs have their force constant as $$k_1$$ and $$k_2\:(k_1 > k_2)$$. When they are stretched by the same force

A
no work is done in case of both the springs.

B
equal work is done in case of both the springs

C
more work is done in case of second spring

D
more work is done in case of first spring.

Solution

From Hooke's law
$$F\propto x\Rightarrow F=kx$$, where k is spring constant

Since force is same in Stretching for both spring so

$$F=k_1x_1 = k_2x_2\Rightarrow x_1<x_2 \: $$ because $$\: k_1>k_2 $$

so work done in case of first spring is $$W_1 =\dfrac{1}{2} k_1x_1^2$$ and work done in case of second spring is

$$\displaystyle W_2 =\dfrac{1}{2}k_2x_2^2 \:so \:\frac{W_1}{W_2}=\frac{x_1}{x_2}\Rightarrow W_1<W_2$$

It means that more work is done in case of second spring (work done on spring is equal to stored elastic potential energy of the spring)
Question 8
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Directions For Questions

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
$$x = 3.00 \,m, y = 3.00 \,m$$.

...view full instructions

Calculate the work done on the tool by $$\vec{F}$$ if this displacement is along the straight line $$y =x$$ that connects these two points.

A
$$2.50 J$$

B
$$500 J$$

C
$$50.6 J$$

D
$$2 J$$

Solution

as , $$dW=f.ds$$

$$W=\int { f.ds } $$

and force is only along the y-axis, work-done along the x-axis is zero

$$\displaystyle W=\int_{0,0}^{3,3} { -\alpha x{ y }^{ 2 }.dy } \cos 180^{\circ}$$ and $$ x= y$$

$$\displaystyle W=\int_{0}^{3} { \alpha { y }^{ 3 }.dy } $$

$$\displaystyle W=\alpha \int _{ 0 }^{ 3 }{ { y }^{ 3 }dy } $$

$$\displaystyle W=2.5*\{ \frac { { 3 }^{ 4 } }{ 4 } -0\} $$

$$W=50.6 J $$
Question 9
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A ball attached to one end of a string swings in a vertical plane such that its acceleration at point A (extreme position) is equal to its acceleration at point B (mean position). The angle $$\displaystyle \theta $$ is

A
$$\displaystyle \cos ^{-1}\left ( \frac{2}{5} \right )$$

B
$$\displaystyle \cos ^{-1}\left ( \frac{4}{5} \right )$$

C
$$\displaystyle \cos ^{-1}\left ( \frac{3}{5} \right )$$

D
None of these

Solution

Let acceleration at both points be 'a'.
At point A, $$mg\sin { \theta } = ma$$
At point B, $$\dfrac { m{ v }^{ 2 } }{ r } = ma$$
By energy conservation, $$\dfrac { m{ v }^{ 2 } }{ 2 } = mgr\times (1-\cos { \theta } )$$
Solving, to get $$\cos { \theta } = -1 \ or \dfrac { 3 }{ 5 }$$
=> θ = $$\cos ^{ -1 }{ \dfrac { 3 }{ 5 } } $$
Question 10
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Directions For Questions

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
$$x = 3.00 \,m, y = 3.00 \,m$$.

...view full instructions

Which of the following statements is correct regarding the work done $$\vec{F}$$ along these two paths.

A
Work done on x-axis is zero

B
Work done on x-axis is less than on y-axis

C
Work done on x-axis is more than on y-axis but not zero

D
Data insufficient

Solution

since force is applied in -ve y-axis and angle between x & y axis is $${ 90 }^{ o }$$
and $$dW=\vec{F}\cdot \vec{ds}$$
$$ W=0 $$ on x-axis
Alternatively, since $$y=0$$ along the path from $$x=0$$( origin) to $$x=3$$, $$\vec{F}=\alpha xy^2 \hat{j}=0$$ $$ \Rightarrow W=\vec{F}\cdot \vec{ds}=0$$
Similarly, $$W=0$$ along y-axis.