Work Energy and Power Test

Result

Work Energy and Power Test - 43

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A particle moves along $$X-$$axis from $$x=0$$ to $$x=1\ m$$ under the influence of a force given by $$F=3x^{2}+2x-10$$. Work done in the process is:

A
$$+4\ J$$

B
$$-4\ J$$

C
$$+8\ J$$

D
$$-8\ J$$

Solution

Given,
$$F=3x^2+2x-10$$
Work done $$=\int { F\cdot ds }$$
$$ =\int _{ 0 }^{ 1 }{ \left( { 3x }^{ 2 }+2x-10 \right) dx } $$
$$=\left[{ \frac { { 3x }^{ 3 } }{ 3 } +\frac { { 2x }^{ 2 } }{ 2 } -10x }\right]_{ 0 }^{ 1 }$$
$$=(1)^3+(1)^2-10-0$$
$$=2-10$$
$$=-8J$$
Question 2
1 / -0

A block of mass $$m$$ at the end of a string is whirled round in a vertical circle of radius $$R$$. The critical speed of the block at the top of its swing below which the string would slacken before the block reaches the top is:

A
$$Rg$$

B
$${(Rg)}^{2}$$

C
$$R/g$$

D
$$\sqrt {Rg}$$

Solution

Correct Option: D

Explanation:
If the block is moving at critical speed, the tension in the string will be zero at the top, and the centripetal force will be provided by gravity:

$$mg=\dfrac{m{{v}^{2}}}{R}~\Rightarrow v=\sqrt{Rg}$$
Question 3
1 / -0

$$4\ J$$ of work is required to stretch a spring through $$10\ cm$$ beyond its unstreched length. The extra work required to stretch it through additional $$10\ cm$$ shall be

A
$$4\ J$$

B
$$8\ J$$

C
$$12\ J$$

D
$$16\ J$$

Solution

W = 4J
X = 10cm = 0.1 m = $$10^{-1}$$
By Hook's law,
W = $$\frac{1}{2}$$k$$x^2$$_________eq1
4 = $$\frac{1}{2}$$ x k x 0.1 x 0.1
8 = k x $$10^{-2}$$
k = 800 N$$m^{-1}$$
Additional 10cm means 0.1 + 0.1 i.e 0.2
Therefore again by same law,
$$w_1$$ = $$\frac{1}{2}$$ x 800 x 4 x $$10^{-2}$$
$$w_1$$ = $$16 J$$
AS we know,
w = $$w_1$$ - W
w = $$12 J$$
Question 4
1 / -0

Under the action of a force, a $$2\ kg$$ body moves such that its position $$x$$ as a function of time $$t$$ is given by $$x=\dfrac{t^{3}}{3}$$, where $$x$$ is in meter and $$t$$ in second. The work done by the force in first two seconds is:

A
$$1600\ J$$

B
$$160\ J$$

C
$$16\ J$$

D
$$\dfrac{16}{9}\ J$$

Solution

Given,
$$x=\dfrac{t^3}{3}$$
Differentiate it wrt t
$$\dfrac{dx}{dt}=\dfrac{3}{3}t^2$$
$$\dfrac{dx}{dt}=t^2$$
$$v=t^2$$
Workdone (W)= Change in K.E
$$W=K.E_f-K.E_i$$
$$K.E_i=0$$
$$K.E_f=\dfrac{1}{2}mv^2$$
$$K.E_f=\dfrac{1}{2}\times 2\times 4^2$$
$$K.E_f=16J$$
$$W=16J$$
Question 5
1 / -0

A particle moves along $$x-$$axis from $$x=0$$ to $$x=5$$ meter under the influence of a force $$F=7-2x+3x^{2}$$. The work done in the process is:

A
$$70$$

B
$$135$$

C
$$270$$

D
$$35$$

Solution

Given,
$$F=3x^2-2x+7$$
Work done $$=\int { F\cdot ds }$$
$$ =\int _{ 0 }^{ 5 }{ \left( { 3x }^{ 2 }-2x+7 \right) dx } $$
$$=\left[{ \frac { { 3x }^{ 3 } }{ 3 } -\frac { { 2x }^{ 2 } }{ 2 } +7x }\right]_{ 0 }^{ 5}$$
$$=(5)^3-(5)^2+7\times 5-0$$
$$=125-25+35$$
$$=135J$$
Question 6
1 / -0

A boy is swinging in a swing. If he stands the time period will

A
First decreases, then increase

B
Decrease

C
Increase

D
Remain same

Solution

As the child stand up the effective length of pendulum decreases due to the reason that center of gravity rises up.
According to $$T=2\pi\sqrt{\left(\dfrac{l}{g}\right)}$$
Thus, the time peroid decreases
Question 7
1 / -0

A body of mass $$4\ kg$$ moves under the action of a force $$\overrightarrow { F } =\left( \hat { 4i } +12{ t }^{ 2 }\hat { j } \right) N$$, where $$t$$ is the time in second. The initial velocity of the particle is $$(2\hat { i } +\hat { j } +2\hat { k } )m{s}^{-1}$$. If the force is applied for $$1\ s$$, work done is:

A
$$4\ J$$

B
$$8\ J$$

C
$$12\ J$$

D
$$16\ J$$

Solution

Given Mass $$m=4kg$$
Force $$\overrightarrow { F } =\left( 4\hat { i } +12{ t }^{ 2 }\hat { j } \right) N$$
Initial velocity $$\overrightarrow { v } =\left( 2\hat { i } +\hat { j } +2\hat { k } \right) m/s$$
Magnitude of $$\overrightarrow { v } =\sqrt { { 2 }^{ 2 }+{ I }^{ 2 }+{ I }^{ 2 } } =\sqrt { 9 } \quad =3m/s$$
Net force $$=\left| \left( 4i+\left| 2{ t }^{ 2 }\hat { j } \right| \right) N \right| \\ \Rightarrow ma=\sqrt { { 16 }^{ 2 }+144{ t }^{ 4 } } \\ \Rightarrow 4a=4\sqrt { 1+9{ t }^{ 4 } } \\ \Rightarrow a=\sqrt { 1+9{ t }^{ 4 } } \\ \Rightarrow \cfrac { dv }{ dt } =\sqrt { 1+9{ t }^{ 4 } } \\ \Rightarrow v=\sqrt { { 1 }^{ 2 }+{ \left( { 3t }^{ 2 } \right) }^{ 2 } } \\ at(t=1)\quad =4$$
Work $$=$$ Force $$\times $$ displacement
$$=12.6\times \cfrac { 1 }{ 3 } =4J$$
Question 8
1 / -0

Under the action of a force, a $$2lg$$ body moves such that it position $$x$$ as a function of time is given by $$x=\cfrac{{t}^{3}}{3}$$, where $$x$$ is in metre and $$t$$ in seconds. The work done by the force in the first two seconds is

A
$$1600J$$

B
$$160J$$

C
$$16J$$

D
$$1.6J$$

Solution

Given : Mass m$$=2kg$$ $$x(t)=\cfrac { { t }^{ 3 } }{ 3 } $$
Time $$t=2s$$
Differentiating position $$x(t)= velocity v$$
$$=\cfrac { d }{ dt } \left( \cfrac { { t }^{ 3 } }{ 3 } \right) =\cfrac { 1 }{ 3 } .3{ t }^{ 2 }\quad ={ t }^{ 2 }$$
At $$t=2s\Rightarrow v={ t }^{ 2 }\quad =4m/s$$
Kinetic energy $$=\cfrac { 1 }{ 2 } m{ v }^{ 2 }\quad =16J$$
Question 9
1 / -0

A particle is moving in a vertical circle the tension in the string when passing through two position at angle $${30}^{o}$$ and $${60}^{o}$$ from vertical lowest position are $${T}_{1}$$ and$${T}_{2}$$ respectively then-

A
$${T}_{1}={T}_{2}$$

B
$${T}_{1}> {T}_{2}$$

C
$${T}_{1}< {T}_{2}$$

D
$${T}_{1}\ge {T}_{2}$$

Solution

$$T-mgcos\theta =\dfrac { m{ v }^{ 2 } }{ r } \\ T=mgcos\theta +\dfrac { m{ v }^{ 2 } }{ r } \\ for\quad \theta =30\\ { T }_{ 1 }=mgcos30+\dfrac { m{ v }^{ 2 } }{ r } \\ for\quad \theta =60\\ { T }_{ 2 }=mgcos60+\dfrac { m{ v }^{ 2 } }{ r } \\ { T }_{ 1 }>{ T }_{ 2 }$$
Question 10
1 / -0

A weightless thread can support tension upto $$30N$$. A particle of mass $$0.5kg$$ is tied to it and is revolved in a circle of radius $$2m$$ in a vertical plane. If $$g=10m/{s}^{2}$$, then the maximum angular velocity of the stone will be

A
$$5rad/s$$

B
$$\sqrt{30}rad/s$$

C
$$\sqrt {60}rad/s$$

D
$$10rad/s$$

Solution

$$T-mg=m\omega ^{ 2 }r\\ { T }_{ max }=mg+m\omega ^{ 2 }r\\ { \omega }^{ 2 }=\left( \frac { T-mg }{ mr } \right) \\ =\left( \frac { 30-5 }{ 0.5(2) } \right) \\ =5rad/sec$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Work Energy and Power Test - 43

Result

Work Energy and Power Test - 43

Score

-

Rank

-

SHARING IS CARING

Question 1

$$+4\ J$$

$$-4\ J$$

$$+8\ J$$

$$-8\ J$$

Solution

Question 2

$$Rg$$

$${(Rg)}^{2}$$

$$R/g$$

$$\sqrt {Rg}$$

Solution

Question 3

$$4\ J$$

$$8\ J$$

$$12\ J$$

$$16\ J$$

Solution

Question 4

$$1600\ J$$

$$160\ J$$

$$16\ J$$

$$\dfrac{16}{9}\ J$$

Solution

Question 5

$$70$$

$$135$$

$$270$$

$$35$$

Solution

Question 6

First decreases, then increase

Decrease

Increase

Remain same

Solution

Question 7

$$4\ J$$

$$8\ J$$

$$12\ J$$

$$16\ J$$

Solution

Question 8

$$1600J$$

$$160J$$

$$16J$$

$$1.6J$$

Solution

Question 9

$${T}_{1}={T}_{2}$$

$${T}_{1}> {T}_{2}$$

$${T}_{1}< {T}_{2}$$

$${T}_{1}\ge {T}_{2}$$

Solution

Question 10

$$5rad/s$$

$$\sqrt{30}rad/s$$

$$\sqrt {60}rad/s$$

$$10rad/s$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.