Work Power & Energy — Page 2 | JEE Physics

Q11

A body of mass

m

is accelerated uniformly from rest to a speed

v

in a time

T.

The instantaneous power delivered to the body as a function of time is given by

A

{{m{v^2}} \over {{T^2}}}.{t^2}

B

{{m{v^2}} \over {{T^2}}}.t

C

{1 \over 2}{{m{v^2}} \over {{T^2}}}.{t^2}

D

{1 \over 2}{{m{v^2}} \over {{T^2}}}.t

Correct Answer

Option B

Solution

u = 0;v = u + aT;v = aT

Instantaneous power

= F \times v = m.\,a.\,at = m.{a^2}.t

$\therefore$ Instantaneous power

= {{m{v^2}t} \over {{T^2}}}

Q12

The upper half of an inclined plane with inclination

\phi

is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is given by

A

2\,\cos \,\,\phi

B

2\,sin\,\,\phi

C

\,\tan \,\,\phi

D

2\,\tan \,\,\phi

Correct Answer

Option D

Solution

Let the length of the inclined plane is =

l

. So only

{l \over 2}

part will have friction. According to work-energy theorem,

W = \Delta k = 0

(Since initial and final speeds are zero) $\therefore$ Work done by friction + Work done by gravity

=0

i.e.,

- \left( {\mu \,mg\,\cos \,\phi } \right){\ell \over 2} + mg\ell \,\sin \,\phi = 0

or

{\mu \over 2}\cos \,\phi = \sin \phi

or

\mu = 2\,\tan \,\phi

Q13

A

2

kg

block slides on a horizontal floor with a speed of

4m/s.

It strikes a uncompressed spring, and compress it till the block is motionless. The kinetic friction force is

15N

and spring constant is

10, 000

N/m.

The spring compresses by

A

8.5cm

B

5.5cm

C

2.5cm

D

11.0cm

Correct Answer

Option B

Solution

Let the block compress the spring by

x

before coming to rest.

Initial kinetic energy of the block $=$ (potential energy of compressed spring) $+$ work done due to friction.

{1 \over 2} \times 2 \times {\left( 4 \right)^2} = {1 \over 2} \times 10000 \times {x^2} + 15 \times x

10,000{x^2} + 30x - 32 = 0

\Rightarrow 5000{x^2} + 15x - 16 = 0

$\therefore$

x = {{ - 15 \pm \sqrt {{{\left( {15} \right)}^2} - 4 \times \left( {5000} \right)\left( { - 16} \right)} } \over {2 \times 5000}}

\,\,\,\,\, = 0.055m = 5.5cm.

Q14

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

A

200J-500J

B

2 \times {10^5}J - 3 \times {10^5}J

C

20,000J - 50,000J

D

2,000J - 5,000J

Correct Answer

Option D

Solution

The average speed of the athelete

v = {{100} \over {10}} = 10m/s\,\,\,\,

$\therefore$

K.E. = {1 \over 2}m{v^2}

If mass of athlete is

40

kg

then,

K.E.

= {1 \over 2} \times 40 \times {\left( {10} \right)^2} = 2000J

If mass of athlete is

100

kg

then,

K.E.

= {1 \over 2} \times 100 \times {\left( {10} \right)^2} = 5000J

His kinetic energy can be in the range = 2000 J to 5000 J.

Q15

A porter lifts a heavy suitcase of mass 80 kg and at the destination lowers it down by a distance of 80 cm with a constant velocity. Calculate the work done by the porter in lowering the suitcase. (take g = 9.8 ms

-

2)

A +627.2 J

B

-

62720.0 J

C

-

627.2 J

D 784.0 J

Correct Answer

Option C

Solution

W = - N \times \Delta x

= - 80 \times 9.8 \times {{80} \over {100}}

= - 627.2

J

Q16

A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 sec. will be:

A 18 J

B 4.5 J

C 22 J

D 9 J

Correct Answer

Option B

Solution

Given that, F = 6t We know, F = ma =

m{{dv} \over {dt}}

$\therefore$

m{{dv} \over {dt}} = 6t

$\Rightarrow$

1.{{dv} \over {dt}} = 6t

[as m = 1] $\Rightarrow$

\int\limits_0^v {dv} = \int {6t} dt

$\Rightarrow$

v = 6\left[ {{{{t^2}} \over 2}} \right]_0^1

$\Rightarrow$

v = {6 \over 2} = 3

m/s [ as given t = 1 sec ] Work done by the body during the first 1 form work-energy theorem, W =

\Delta

K.E =

{1 \over 2}m\left( {{V^2} - {v^2}} \right)

=

{1 \over 2}.1.\left( {{3^2} - {0^2}} \right)

= 4.5 J

Q17

A uniform cable of mass 'M' and length 'L' is placed on a horizontal surface such that its (1/n)th part is hanging below the edge of the surface. To lift the hanging part of the cable upto the surface, the work done should be :

A

{{2MgL} \over {{n^2}}}

B nMgL

C

{{MgL} \over {2{n^2}}}

D

{{MgL} \over {{n^2}}}

Correct Answer

Option C

Solution

To solve this problem, we need to determine the work done to lift the hanging part of the cable up to the surface.

The work done lifting a small element of the cable will be the weight of the element times the distance it has to be lifted.

Let's take a small section of the cable at a depth $x$ below the surface.

This section has a length of $dx$ , so its mass is $(M/L)dx$ where $(M/L)$ is the linear mass density of the cable.

The work $dW$ done to lift this small section up to the surface is the weight of the section times the distance it has to be lifted : $dW = (M/L)gdx \times x$ .

Integrating this expression from 0 to L/n (the length of the hanging part of the cable) gives the total work done :

W = \int\limits_{0}^{L/n} (M/L)gxdx

= (Mg/L) \int\limits_{0}^{L/n} xdx

= (Mg/L) \times [x^2/2]_{0}^{L/n}

= (Mg/L) \times [L^2/(2n^2)]

= MgL/(2n^2)

So the work done to lift the hanging part of the cable up to the surface is $MgL/(2n^2)$ .

Therefore, the correct answer is Option C :

\frac{MgL}{2n^2}

Q18

If the potential energy between two molecules is given by U =

- {A \over {{r^6}}} + {B \over {{r^{12}}}}

, then at equilibrium, separation between molecules, and the potential energy are :

A

{\left( {{{2B} \over A}} \right)^{1/6}}

,

- {{{A^2}} \over {4B}}

B

{\left( {{{2B} \over A}} \right)^{1/6}}, - {{{A^2}} \over {2B}}

C

{\left( {{B \over A}} \right)^{1/6}},0

D

{\left( {{B \over {2A}}} \right)^{1/6}}, - {{{A^2}} \over {2B}}

Correct Answer

Option A

Solution

U =

- {A \over {{r^6}}} + {B \over {{r^{12}}}}

F = -

{{dU} \over {dr}}

= – (A(–6r–7 )) + B(–12r–13) for equilibrium, F = 0 $\therefore$ 0 =

{{6A} \over {{r^7}}} - {{12B} \over {{r^{13}}}}

$\Rightarrow$

{{6A} \over {12B}} = {1 \over {{r^6}}}

$\Rightarrow$ r =

{\left( {{{2B} \over A}} \right)^{{1 \over 6}}}

$\therefore$ U =

- {A \over {{{2B} \over A}}} + {B \over {{{\left( {{{2B} \over A}} \right)}^2}}}

=

- {{{A^2}} \over {2B}} + {{{A^2}} \over {4B}}

=

- {{{A^2}} \over {4B}}

Q19

A person pushes a box on a rough horizontal plateform surface. He applies a force of 200 N over a distance of 15 m. Thereafter, he gets progressively tired and his applied force reduces linearly with distance to 100 N. The total distance through which the box has been moved is 30 m. What is the work done by the person during the total movement of the box?

A 5690 J

B 5250 J

C 2780 J

D 3280 J

Correct Answer

Option B

Solution

Work done = area of ABCEO = area of trapezium ABCD + area of rectangle ODCE =

{1 \over 2}

$\times$ 45 $\times$ 30 + 100 $\times$ 30 = 5250J

Q20

A 60 HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg. If the frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to : (1 HP = 746 W, g = 10 ms-2)

A 1.5 ms-1

B 1.7 ms-1

C 2.0 ms-1

D 1.9 ms-1

Correct Answer

Option D

Solution

F = mg + f F = 20000 + 4000 = 24000 N We know, Power(P) = Fv $\Rightarrow$ v =

{P \over F}

=

{{60 \times 746} \over {24000}}

$\Rightarrow$ v $\approx$ 1.9 m/s