Work Energy and Power Test

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Work Energy and Power Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

A particle moves under the effect of a force $$F=Cx$$ from $$x=0$$ to $$x={x}_{1}$$. The work done in the process is

A
$$C{x}_{1}^{2}$$

B
$$\cfrac{1}{2} C{x}_{1}^{2}$$

C
$$C{x}_{1}$$

D
Zero

Solution

Here F = Cx
Work done = F.d Where d is displacement

$$w = \int^{x_2}_{x_1} F.dx$$

$$w = \int^{x_1}_{0} Fdx$$

$$w = \int^{x_1}_{0} cx \space dx$$
$$w = \dfrac{1}{2} cx^2_1$$
Here (B) is correct answer
Question 2
1 / -0

A force acts on a $$3\ g$$ particle in such a way that the position of the particle as a function of time is given by $$x=3t-4t^{2}+t^{3}$$, where $$x$$ is in meters and $$t$$ is in second. The work done during the first $$4$$ second is:

A
$$490\ mJ$$

B
$$450\ mJ$$

C
$$528\ mJ$$

D
$$530\ mJ$$

Solution

Given,
$$ x=3t-4{{t}^{2}}+{{t}^{3}} $$
$$ dx=d\left( 3t-4{{t}^{2}}+{{t}^{3}} \right)=\left( 3-8t+3{{t}^{2}} \right)dt $$
Acceleration
$$a(x)=\dfrac{{{d}^{2}}x}{dt}=\dfrac{{{d}^{2}}x}{dt}=\dfrac{d\left( 3-8t+3{{t}^{2}} \right)}{dt}=-8+6t$$
Work done =$$dw=ma.dx=m\left( 6t-8 \right)\left( 3-8t+3{{t}^{2}} \right)dt$$
$$ \int_{0}^{W}{dw}=m\int_{0}^{4}{\left( 6t-8 \right)\left( 3-8t+3{{t}^{2}} \right)dt} $$
$$ W=m\times 176=0.003\times 176=528\times {{10}^{-3}}\,J $$

$$ W=528\times {{10}^{-3}} J $$
Question 3
1 / -0

Two particles of masses $${m}_{1}$$ and $${m}_{2}$$ in projectile motion have velocities $${v}_{1}$$ and $${v}_{2}$$ respectively at time $$t=0$$. They collide at time $${t}_{0}$$. Their velocities become $${v'}_{1}$$ and $${v'}_{2}$$ at time $$2{t}_{0}$$ while still moving in air. The value of $$\left[ \left( { m }_{ 1 }{ v' }_{ 1 }+{ m }_{ 2 }{ v' }_{ 2 } \right) -\left( { m }_{ 1 }{ v }_{ 1 }+{ m }_{ 2 }{ v }_{ 2 } \right) \right] $$

A
zero

B
$$({m}_{1}+{m}_{2})g{t}_{0}$$

C
$$2({m}_{1}+{m}_{2})g{t}_{0}$$

D
$$\cfrac{1}{2}({m}_{1}+{m}_{2})g{t}_{0}$$

Solution

External force = $$F_{ext}=\dfrac{\Delta p}{\Delta t}=((m_{1}v_{1}^{'}+m_{2}v_{2}^{'})-(m_{1}v_{1}+m_{2}v_{2}))/2t_{0}-0$$
$$((m_{1}v_{1}^{'}+m_{2}v_{2}^{'})-(m_{1}v_{1}+m_{2}v_{2})=2t_{0}F_{ext}=2to(m_{1}+m_{2})g$$
Question 4
1 / -0

A stone of mass $$0.2kg$$ is tied to one end of a thread of length $$0.1m$$ whirled in a vertical circle. When the stone is at the lowest point of circle, tension in thread is $$52N$$, then velocity of the stone will be:

A
$$4m/s$$

B
$$5m/s$$

C
$$6m/s$$

D
$$7m/s$$

Solution

$$T-mg=\dfrac { { mv }^{ 2 } }{ r } \\ T=\dfrac { { mv }^{ 2 } }{ r } +mg\\ 52=0.2(10)\quad +\dfrac { 0.2({ v }^{ 2 }) }{ 0.1 } \\ 2{ v }^{ 2 }=52-2=50\\ \qquad v=5m/sec$$
Question 5
1 / -0

A stone of mass $$1kg$$ is tied to the end of a string of $$1m$$ length. It is whirled in a vertical circle. If the velocity of the stone at the top be $$4m/s$$. What is the tension in the string (at that instant)?

A
$$6N$$

B
$$16N$$

C
$$5N$$

D
$$10N$$

Solution

$$T+mg=\dfrac { { mv }^{ 2 } }{ r } \\ T=\dfrac { { mv }^{ 2 } }{ r } -mg\\ \dfrac { 1\times { 4 }^{ 2 } }{ 1 } -1(10)\\ =16-10\\ =6N$$
Question 6
1 / -0

Force constant of spring one is $$K$$ and another spring is $$2K$$. When both spring are stretched through same distance, then the work done

A
$${W}_{2}=2{W}_{1}^{2}$$

B
$${W}_{2}=2{W}_{1}$$

C
$${W}_{2}={W}_{1}$$

D
$${W}_{2}=0.5{W}_{1}$$

Solution

Work done = - $$\Delta (Potential energy)$$
= $$-\frac{1}{2}kx^{2}$$
Both springs are stretched by the same length , let it be $$l$$ .
Then ,
$$W_{1}$$ = $$-\dfrac{1}{2}kl^{2}$$
$$W_{2}$$ = $$-\dfrac{1}{2}2kl^{2}$$
Therefore , $$W_{2}$$ = $$2$$$$W_{1}$$
Question 7
1 / -0

A particle of mass $$m$$ initially moving with speed $$v$$. A force acts on the particle $$f=kx$$ where $$x$$ is the distance travelled by the particle and $$k$$ is constant. Find the speed of the particle when the work done by the force equals $$W$$.

A
$$\sqrt { \frac { k }{ m } +{ v }^{ 2 } }$$

B
$$\sqrt { \frac { 2W }{ m } +{ v }^{ 2 } }$$

C
$$\sqrt { \frac { 2W }{ k } +{ v }^{ 2 } }$$

D
$$\sqrt { \frac { W }{ 2m } +{ v }^{ 2 } }$$

Solution

Given,

Final velocity, $${{v}_{1}}$$

Change in kinetic Energy = work done by force

$$ \dfrac{1}{2}mv_{1}^{2}-\dfrac{1}{2}m{{v}^{2}}=W $$

$$ {{v}_{1}}=\sqrt{\dfrac{2W}{m}+{{v}^{2}}} $$

Hence, final velocity is $$\sqrt{\dfrac{2W}{m}+{{v}^{2}}}$$
Question 8
1 / -0

A wheel rotating at an angular speed of 20 rad$${ s }^{ -1 }$$, is brought to rest by a constant torque in 4s.If the M.I is 0.2 kg $${ m }^{ 2 }$$, the work done in first 2s is:

A
$$50\ J$$

B
$$30\ J$$

C
$$20\ J$$

D
$$10\ J$$

Solution

Angular acceleration$$\alpha=\dfrac{20}{4}=5$$
Torque=$$I\alpha=5\times 0.2=1$$
Angular displacement in first 2 secs=$$20\times 2+\dfrac{1}{2}5\times 2^2=50$$
Work done=Torque×angular displacement$$=1×50=50J$$
Question 9
1 / -0

A force of $$(4{ x }^{ 2 }+3x)N$$ acts on a particle which displaces it from $$x=2m$$ to $$x=3m$$. The work done by the force is:

A
$$32.8 J$$

B
$$3.28 J$$

C
$$0.328 J$$

D
$$zero$$

Solution

$$W=\int Fdx=\int (4x^2+3x)dx=\dfrac{4x^3}{3}+\dfrac{3x^2}{2}$$
$$W=\dfrac{4\times 3^3}{3}+\dfrac{3\times 3^2}{2}-\dfrac{4\times 2^3}{3}-\dfrac{3\times 2^2}{2}=32.8J$$
Question 10
1 / -0

A force of $$(4x^{2}+3x)\ N$$ acts on a particle which displaces it from $$x=2m$$ to $$x=3m$$. Te work done by the force is:

A
$$32.8\ J$$

B
$$3.281\ J$$

C
$$0.328\ J$$

D
$$zero$$

Solution

$$W=F dx=\int (4x^2+3x)dx=\dfrac{4x^3}{3}+\dfrac{3x^2}{2}=32.8J$$

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Re-Attempt Weekly Quiz Competition

Work Energy and Power Test - 44

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Work Energy and Power Test - 44

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SHARING IS CARING

Question 1

$$C{x}_{1}^{2}$$

$$\cfrac{1}{2} C{x}_{1}^{2}$$

$$C{x}_{1}$$

Zero

Solution

Question 2

$$490\ mJ$$

$$450\ mJ$$

$$528\ mJ$$

$$530\ mJ$$

Solution

Question 3

zero

$$({m}_{1}+{m}_{2})g{t}_{0}$$

$$2({m}_{1}+{m}_{2})g{t}_{0}$$

$$\cfrac{1}{2}({m}_{1}+{m}_{2})g{t}_{0}$$

Solution

Question 4

$$4m/s$$

$$5m/s$$

$$6m/s$$

$$7m/s$$

Solution

Question 5

$$6N$$

$$16N$$

$$5N$$

$$10N$$

Solution

Question 6

$${W}_{2}=2{W}_{1}^{2}$$

$${W}_{2}=2{W}_{1}$$

$${W}_{2}={W}_{1}$$

$${W}_{2}=0.5{W}_{1}$$

Solution

Question 7

$$\sqrt { \frac { k }{ m } +{ v }^{ 2 } }$$

$$\sqrt { \frac { 2W }{ m } +{ v }^{ 2 } }$$

$$\sqrt { \frac { 2W }{ k } +{ v }^{ 2 } }$$

$$\sqrt { \frac { W }{ 2m } +{ v }^{ 2 } }$$

Solution

Question 8

$$50\ J$$

$$30\ J$$

$$20\ J$$

$$10\ J$$

Solution

Question 9

$$32.8 J$$

$$3.28 J$$

$$0.328 J$$

$$zero$$

Solution

Question 10

$$32.8\ J$$

$$3.281\ J$$

$$0.328\ J$$

$$zero$$

Solution

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