Work Energy and Power Test

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Work Energy and Power Test - 67

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Question 1
1 / -0

A small sphere is given vertical velocity of magnitude $$v_{0} = 5m/s$$ and it swings in a vertical plane about the end of massless string. The angle $$\theta$$ with the vertical at which string will break, knowing that it can withstand a maximum tension equal to twice the weight of the sphere, is $$[g = 10 m/s^{2}]$$.

A
$$\cos^{-1}\left (\dfrac {2}{3}\right )$$

B
$$\cos^{-1}\left (\dfrac {1}{4}\right )$$

C
$$60^{\circ}$$

D
$$30^{\circ}$$

Solution
Question 2
1 / -0

Work done in time $$t$$ on a body of mass $$m$$ which is accelerated from rest to a speed $$v$$ in time as a function of time $$t$$ is given by

A
$$\cfrac { 1 }{ 2 } m\cfrac { v }{ { t }_{ 1 } } { t }^{ 2 }\quad $$

B
$$m\cfrac { v }{ { t }_{ 1 } } { t }^{ 2 }$$

C
$$\cfrac { 1 }{ 2 } m{ \left( \cfrac { v }{ { t }_{ 1 } } \right) }^{ 2 }{ t }^{ 2 }$$

D
$$\cfrac { 1 }{ 2 } m\cfrac { v }{ { t }_{ 1 }^{ 2 } } { t }^{ 2 }$$

Solution

Ans : We know $$ W = F \times s $$
where,
$$ F = ma $$
By using Kinematics, $$ s = \dfrac{1}{2}at^{2} $$ , if u =0

So,
$$ W = (ma).(\dfrac{1}{2}at^{2}) $$
$$ W = \dfrac{1}{2}ma^{2}t^{2} $$ where $$ a = v/t_{1} $$

$$ \boxed{W = \dfrac{1}{2}m(\dfrac{v}{t_{1}})^{2}t^{2}} $$
Question 3
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A mass tied to a string moves in a vertical circle and at the point $$P$$ its speed is $$5m/s$$. At the point $$P$$ the string breaks. The mass will reaches height above $$P$$ of nearly $$\left( g=10m/{ s }^{ 2 } \right) $$

A
$$1m$$

B
$$0.5m$$

C
$$1.27m$$

D
$$1.25m$$

Solution
Question 4
1 / -0

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

A
$$10\ m/s$$

B
$$5\sqrt{2}\ m/s$$

C
$$10\sqrt{3}\ m/s$$

D
$$20\ m/s$$

Solution

$$The\quad ratio\quad to\quad the\quad mamimum\quad { T }_{ 2 }{ \quad to\quad the\quad minimum\quad tension\quad T }_{ 1 }\quad is\quad -\\ \frac { { T }_{ 2 } }{ { T }_{ 1 } } =4\\ { T }_{ 2 }=4{ T }_{ 1 }\\ Now\quad the\quad difference\quad between\quad the\quad two\quad 6mg;\\ Hence\quad we\quad have\quad \\ { T }_{ 2 }-{ T }_{ 1 }=6mg\\ \therefore 4{ T }_{ 1 }-{ T }_{ 1 }=6mg\\ 3{ T }_{ 1 }=6mg\\ { T }_{ 1 }=2mg\\ Now\quad tension\quad at\quad the\quad top\quad of\quad the\quad circle\quad is-\\ { T }_{ 1+mg }=\frac { { mv }_{ 1 }^{ 2 } }{ r } =\frac { { mv }_{ 1 }^{ 2 } }{ L } \\ 2mg+mg=\frac { { mv }_{ 1 }^{ 2 } }{ \frac { 10 }{ 3 } } \\ { v }_{ 1 }^{ 2 }=3g\frac { 10 }{ 3 } =10g\\ { v }_{ 1 }=\sqrt { 10g } \\ =\sqrt { 10\times 10 } =10m/s$$
Question 5
1 / -0

From $$x=0$$ to $$x=6$$, the force experienced by an object varies according to the function $$F\left(x\right)=\sqrt{6x-{x}^{2}}$$, as shown above. What is the work done by this force as the object moves from $$x=0$$ to $$x=3$$?
Assume all numbers are given in standard units.

A
$$0 J$$

B
$$4.5 J$$

C
$$7.1 J$$

D
$$9.0 J$$

E
$$14.0 J$$

Solution

There is a simpler method to arrive at the solution.
Conceptually, you are integrating force over displacement, so if you are given the graph, the integration is equal to the area under the curve. The plot given is one of a semicircle.
The function $$F\left( x \right) =\sqrt { 6x-{ x }^{ 2 } }$$ is one of a semicircle.
It may be easier to see this when it is in the form $$F(x)=\sqrt{9+6x-x^2-9}$$ , which corresponds to $$ F(x)=3^2-(x-3)^2$$ which is the upper half of a circle centered at $$x = 3$$ with radius $$3$$.
The area of a circle is given by $$A=\pi r^2$$, and in this case you are looking for the area under half of the top half.
Thus, $$W=\dfrac{1}{4}A=\dfrac{1}{4}\pi (3)^2\approx7.0658\ J$$, that is choice "C".
Question 6
1 / -0

An athlete in the Olympic games covers a distance of $$100$$m in $$10$$s. His kinetic energy can be estimated to be in the range. (Assume weight = 60kg)

A
$$200J-500J$$

B
$$2\times 10^5J-3\times 10^5J$$

C
$$20000J-50000J$$

D
$$2000J-5000J$$

Solution

Velocity $$V=\dfrac{distance}{time}=\dfrac{100}{10}=10 m/s$$

Assuming his mass to be 60kg, his kinetic energy is
$$K.E =\dfrac{1}{2}mv^2 $$

$$= \dfrac{1}{2}\times 60\times 100$$

$$=60\times50$$

$$=3000 J$$
Hence, Option D
Question 7
1 / -0

Two blocks are connected by a massless string that passes over a frictionless peg as shown in fig. One end of the string is attached to a mass $$m_1 = 3kg$$, i.e. a distance $$R = 1.20 m$$ from the peg. The other end of the string is connected to a block of mass $$m_2 = 6 kg$$ resting on a table. When the 3 kg block be released at $$\displaystyle \theta = \frac{\pi}{k}$$, the 6 kg block just lift off the table? Find the value of k.

A
$$2$$

B
$$3$$

C
$$4$$

D
$$6$$

Solution

Tension in the string will be maximum when $$m_1$$ is at lowest position. For $$m_2$$ to just lift, tension in the string at this position should equal the weight of $$m_2$$ at this position.

Using conservation of energy,
$$\dfrac{m_1 R^2 \omega^2}{2} = m_1 g R (1-\cos \theta) $$
$$m_1 \omega^2 R = 2m_1 g (1 - \cos \theta) \quad ........(i) $$

From the FBD of $$m_1$$,
$$T_1 - m_1g = m_1 \omega^2 R \quad ...........(ii)$$

From $$(i)\ \& \ (ii),$$
$$T_1 = m_1g + 2m_1g (1-\cos \theta)$$
$$T_1 = m_1g (3-2\cos\theta) \quad .........(iii)$$

From the FBD of $$m_2$$,
$$T_2 + N_2 = m_2 g \quad ........(iv)$$
When the block just leaves the surface,
$$N_2 = 0 \quad .....(v)$$
Also, tension throughout the string is constant,
$$T_2=T_1 \quad ........(vi) $$

Using $$(iii), \ (iv), \ (v) \ \& \ (vi),$$
$$m_2g = m_1g(3-2cos\theta)$$
$$\cos \theta = \dfrac{3 - m_2/m_1}{2}$$
Substituting values,
$$\cos \theta = 1/2$$
$$\theta = \pi/6$$

$$\implies k = 6$$
Question 8
1 / -0

A car of mass $$m$$ starts moving so that its velocity varies according to the law $$v=\beta \sqrt { s }$$, where $$\beta$$ is a constant, and $$s$$ is the distance covered. The total work performed by all the forces which are acting on the car during the first $$t$$ seconds after the beginning of motion is:

A
$${ m\beta }^{ 4 }{ t }^{ 2 }/8$$

B
$${ m\beta }^{ 2 }{ t }^{ 4 }/8$$

C
$${ m\beta }^{ 4 }{ t }^{ 2 }/4$$

D
$${ m\beta }^{ 2 }{ t }^{ 4 }/4$$

Solution

$$v=\beta \sqrt s$$
$$\dfrac{ds}{dt}=\beta \sqrt s$$
On integration
$$2\sqrt s=\beta t$$
$$s=\dfrac{\beta^2 t^2}{4}$$
$$v=\dfrac{\beta^2 t}{2}$$
On differentiation with respect to t
$$a=\dfrac{\beta^2}{2}$$
Work =F.s=mas=$$\dfrac{m\beta^4 t^2}{8}$$
Question 9
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By applying a force $$\vec{F} = (3xy - 5z)\hat{j} + 4z\hat{k}$$ a particle is moved along the path $$y=x^{2}$$ from point $$(0,0,0)$$ to point $$(2,4,0)$$. The work done by the $$F$$ on the particle is

A
$$\dfrac{280}{5}$$

B
$$\dfrac{140}{5}$$

C
$$\dfrac{232}{5}$$

D
$$\dfrac{192}{5}$$

Solution
Question 10
1 / -0

A small block of mass m is pushed on a smooth track from position A with a velocity $$2\sqrt5$$ times the minimum velocity required to reach point D. The block will leave the contact with track at the point where normal force between them becomes zero.
At what angle $$\theta$$ with horizontal does the block gets separated from the track?

A
$$sin^{-1}(\frac{1}{3})$$

B
$$sin^{-1}(\frac{3}{4})$$

C
$$sin^{-1}(\frac{2}{3})$$

D
never leaves contact with the track

Solution