Work Energy and Power Test

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Work Energy and Power Test - 41

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Question 1
1 / -0

Two wires $$AC$$ and $$BC$$ are tied at $$C$$ of a small sphere of mass $$5\ kg$$, which revolves at a constant speed $$v$$ in the horizontal plane with the speed $$v$$ of radius $$1.6\ m$$. Find the minimum value of $$v$$.

A
$$4\ m{s}^{-1}$$

B
$$2\ m{s}^{-1}$$

C
$$2.5\ m{s}^{-1}$$

D
None of these

Solution

From $$FBD$$
$$T_1\cos 30^0+T_2\cos 45^0=mg$$
$$T_1\sin 30^0+T_2\sin 45^0=\dfrac{mv^2}{r}$$
$$T_1\sin 30^0-T_1\cos 30^0=\dfrac{mv^2}{R}-mg$$ $$\because\{\sin45^0=\cos 45^0\}$$
$$T_1=\dfrac{\dfrac{mv^2}{R}-mg}{\left[\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}\right]}$$

as it a string tension, So it can't be negative
$$T_1 > 0$$ i.e $$\dfrac{mv^2}{R}-mg \ge 0$$
$$\dfrac{v_2}{r}-g \ge 0$$
$$v\ge \sqrt{gR}$$
$$V_{min}\sqrt{gr}=\sqrt{10\times 1.6}=\sqrt{16}=4ms^{-1}$$
Hence (A) is the correct options.
Question 2
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A force F is related to the position of a particle by the relation $$ F =(10x^2)N$$. The work done by the force when the particle moves from x=2m to x=4 m is

A
$$\dfrac{56}{3}J$$

B
$$560 J$$

C
$$\dfrac {560}{3}J$$

D
$$\dfrac{3}{560} J$$

Solution

$$F=10 x^{2}N$$
We know that work done $$=\displaystyle\int_{x_{1}}^{x_{2}}F.dx$$
$$\omega=\displaystyle\int_{2}^{4}10 x^{2}dx$$

$$\Rightarrow \omega=10\displaystyle\int_{2}^{4}x^{2}dx$$

$$=10\left[\dfrac{x^{3}}{3}\right]_{2}^{4}$$

$$=10\left[\dfrac{64}{3}-\dfrac{8}{3}\right]=\dfrac{560}{3}\ J$$
Question 3
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Which of the following is not an example of potential energy?

A
A vibrating pendulum at its maximum displacement from the mean position

B
A body at rest at some height from the ground

C
A wound clock-spring

D
A vibrating pendulum when it is just passing through the mean position

Solution

Whenever a body is at some height from the mean (base) position or at a state of extension (or contraction), the body will have some potential energy. But when the pendulum is passing the mean position, the height from the base is 0. Hence a vibrating pendulum at the mean position has no potential energy.
Question 4
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The velocity of a body moving in a vertical circle of radius r is $$\sqrt{7gr}$$ at the lowest point of the circle. What is the ratio of maximum and minimum tension?

A
$$4: 1$$

B
$$\sqrt 7:1$$

C
$$3: 1$$

D
$$2: 1$$

Solution

Since $$V_{b} >\sqrt{5 gr}\rightarrow$$ It will complete vertical circular
$$\downarrow$$
velocity at bottom
Thus $$T_{bottom}-T_{top}=6\ mg$$
$$T_{max}-T_{min}=6\ mg ...... (1)$$
Now,
$$\Rightarrow T_{max}-mg=m(a_{c})$$
$$T_{max}-mg =\dfrac{mv^{2}}{r}$$
$$T_{max}=\dfrac{m}{r}(7 gr)+mg$$
$$T_{max}=8\ mg$$
From $$(1)$$
$$T_{min}=2\ mg$$
Thus $$\dfrac{T_{max}}{T_{min}}=4:1$$
Question 5
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A body of mass $$0.5\ kg$$ travels in a straight line with velocity $$v = ax^{3/2}$$ where $$a = 5\ m^{-1/2}s^{-1}$$. The work done by the net force during its displacement from $$x = 0$$ to $$x = 2m$$ is

A
$$1.5\ J$$

B
$$50\ J$$

C
$$10\ J$$

D
$$100\ J$$

Solution

Given: $$m = 0.5\ kg, v = kx^{3/2}$$ where, $$k = 5\ m^{-1/2}s^{-1}$$
Acceleration, $$a = \dfrac {dv}{dt} = \dfrac {dv}{dx}\dfrac {dx}{dt} = v\dfrac {dv}{dx} \left (\because v = \dfrac {dx}{dt}\right )$$
As $$v^{2} = k^{2}x^{3}$$
Diferentiating both sides with respect to $$x$$, we get
$$2v \dfrac {dv}{dx} = 3k^{2}x^{2}$$
$$\therefore Acceleration, a = \dfrac {3}{2} k^{2}x^{2}$$
Force, $$F = Mass\times Acceleration = \dfrac {3}{2} mk^{2}x^{2}$$
Work done, $$W = \int Fdx = \int_{0}^{2}\dfrac {3}{2} mk^{2}x^{2}dx$$
$$W = \dfrac {3}{2} mk^{2}\left [\dfrac {x^{3}}{3}\right ]_{0}^{2} = \dfrac {3}{6} \times 0.5\times 5^{2} \times [2^{3} - 0] = 50\ J$$.
Question 6
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A force $$F$$ acting on an object varies with distance $$x$$ as shown in the figure. The work done by the force in moving the object from $$x = 0$$ and $$x = 20\ m$$ is

A
$$500\ J$$

B
$$1000\ J$$

C
$$1500\ J$$

D
$$2000\ J$$

Solution

Area under force displacement curve is the work done in that interval
Area under the given figure$$=$$ Area of surface$$+$$ Area of triangle.
Work done$$=10\times 100+\cfrac { 1 }{ 2 } \times 10\times 100$$
$$=1000+500\\ =1500J$$
Question 7
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A time-varying force $$F=6t-2{ t }^{ 2 } N$$, at $$t=0$$ starts acting on a body of mass $$2 kg$$ initially at rest, where t is in second. The force is withdrawn just at the instant when the body comes to rest again. We can see that at $$t=0$$,the force $$F=0$$. Now answer the following:
Mark the correct statement:

A
Velocity of the body is maximum when force acting on the body is maximum for the first time.

B
The velocity of the body becomes maximum when force acting on the body becomes zero again

C
When force becomes zero again, velocity of the body also becomes zero at that instant.

D
All of the avove

Solution
Question 8
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A man revolves a stone of mass m tied to the end of a string in a vertical circle of radius R, The net force at the lowest and height points of the circle directed vertical downwards are
Here $$T_1, T_2$$ and $$v_1, v_2$$ denote the tension in the string and the speed of the stone at the lowest and highest points, respectively.

A
Lowest point: $$mg - T_1$$ Highest point : $$mg + T_2$$

B
Lowest point: $$mg + T_1$$ Highest point : $$mg - T_2$$

C
Lowest point: $$mg + T_1 -\frac{mv^2}{R}$$ Highest point: $$mg - T_2 +\frac{mv^2}{r}$$

D
Lowest point: $$mg - T_1 -\frac{mv^2}{R}$$ Highest point: $$mg + T +\frac{mv^2}{r}$$

Solution
Question 9
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A force $$F = -K(x\hat {i} + y\hat {j})$$ (where $$K$$ is a positive constant) acts on a particle moving in the $$x-y$$ plane. Starting from the origin, the particle is taken along the positive $$x-$$ axis to the point $$(a, 0)$$ and then to the point $$(a, a)$$. The total work done by the force $$\vec {F}$$ on the particle is

A
$$-2Ka^{2}$$

B
$$2Ka^{2}$$

C
$$-Ka^{2}$$

D
$$Ka^{2}$$

Solution

For motion of the particle from $$(0,0)$$ to $$(a,0)$$
$$\vec{F}=-K(0\hat{i}+a\hat{j})\Rightarrow \vec{F}=-Ka\hat{j}$$
Displacement $$\vec{r}=(a\hat{i}+0\hat{j})-(0\hat{i}+0\hat{j})=a\hat{i}$$
So work done from $$(0,0)$$ to $$(a,0)$$ is given by
$$W=\vec{f}.\vec{r}=-Ka\hat{j}.a\hat{i}=0$$
For motion $$(a,0)$$ to $$(a,a)$$
$$\vec{F}=-K(a\hat{i}+a\hat{j})$$ and displacement
$$\vec{r}=(a\hat{i}+a\hat{j})-(a\hat{i}+0\hat{j})=a\hat{j}$$
So work done from $$(a,0)$$ to $$(a,a)$$ $$W=\vec{F}.\vec{r}$$
$$=-K(a\hat{i}+a\hat{j}).a\hat{j}=-Ka^{2}$$
So total work done $$=-Ka^{2}$$
Question 10
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A spring of spring constant $$5\times {10}^{3}N/m$$ is stretched initially by $$5cm$$ from unstretched position. Then the work required to stretch is further by another $$5cm$$ is then

A
$$12.50N-m$$

B
$$18.75N-m$$

C
$$25.00N-m$$

D
$$6.25N-m$$

Solution

$$\Delta W=\cfrac { 1 }{ 2 } \left( 5\times { 10 }^{ 3 } \right) \left[ { (0.1) }^{ 2 }-{ (0.5) }^{ 2 } \right] =18.75$$