Practice Start Test

Work Power & Energy

NEET Physics · 97 questions · Page 8 of 10 · Click an option or "Show Solution" to reveal answer

Q71
An automobile of mass 'm' accelerates starting from origin and initially at rest, while the engine supplies constant power P. The position is given as a function of time by :
A (9P8m)12t32{\left( {{{9P} \over {8m}}} \right)^{{1 \over 2}}}{t^{{3 \over 2}}}
B (8P9m)12t23{\left( {{{8P} \over {9m}}} \right)^{{1 \over 2}}}{t^{{2 \over 3}}}
C (9m8P)12t32{\left( {{{9m} \over {8P}}} \right)^{{1 \over 2}}}{t^{{3 \over 2}}}
D (8P9m)12t32{\left( {{{8P} \over {9m}}} \right)^{{1 \over 2}}}{t^{{3 \over 2}}}
Correct Answer
Option D
Solution

P = const.

P=Fv=mv2dvdxP = Fv = {{m{v^2}dv} \over {dx}}
∫0xPmdx=∫0vv2dv\int\limits_0^x {{P \over m}dx} = \int\limits_0^v {{v^2}dv}
Pxm=v33{{Px} \over m} = {{{v^3}} \over 3}
(3Pxm)1/3=v=dxdt{\left( {{{3Px} \over m}} \right)^{1/3}} = v = {{dx} \over {dt}}
(3Pm)1/3∫0tdt=∫0xx−1/3dx{\left( {{{3P} \over m}} \right)^{1/3}}\int\limits_0^t {dt} = \int\limits_0^x {{x^{ - 1/3}}} dx
⇒x=(8P9m)1/2t3/2\Rightarrow x = {\left( {{{8P} \over {9m}}} \right)^{1/2}}{t^{3/2}}
Q72
If a body looses half of its velocity on penetrating 33 cmcm in a wooden block, then how much will it penetrate more before coming to rest?
A 11 cmcm
B 22 cmcm
C 33 cmcm
D 44 cmcm
Correct Answer
Option A
Solution

We know the work energy theorem,

W=ΔK=FSW = \Delta K = FS

For first penetration, by applying work energy theorem we get,

12mv2−12m(v2)2=F×3  ...(i){1 \over 2}m{v^2} - {1 \over 2}m{\left( {{v \over 2}} \right)^2} = F \times 3\,\,...(i)

For second penetration, by applying work energy theorem we get,

12m(v2)2−0=F×S ...(ii){1 \over 2}m{\left( {{v \over 2}} \right)^2} - 0 = F \times S\,...(ii)

On dividing

(ii)(ii)

by

(i)(i)
1/43/4=S/3{{1/4} \over {3/4}} = S/3

∴\therefore

S=1 cmS = 1\,cm
Q73
A particle is moving in a circular path of radius aa under the action of an attractive potential U=−k2r2U = - {k \over {2{r^2}}} Its total energy is:
A −32ka2 - {3 \over 2}{k \over {{a^2}}}
B Zero
C −k4a2 - {k \over {4{a^2}}}
D k2a2 {k \over {2{a^2}}}
Correct Answer
Option B
Solution

We know, Total energy = Kinetic energy + Potential energy Potential energy given as

U=−k2r2U = - {k \over {2{r^2}}}

We need to find Kinetic Energy. As Force acting on the particle (F) =

−dUdr- {{dU} \over {dr}}
⇒F=−ddr(−k2r2)\Rightarrow F = - {d \over {dr}}\left( {{{ - k} \over {2{r^2}}}} \right)
=k2×(−2)×r−3= {k \over 2} \times \left( { - 2} \right) \times {r^{ - 3}}
=−kr3= - {k \over {{r^3}}}

Because of this force particle is having circular motion so it will provide possible centripetal force.

∣F∣=mv2r\left| F \right| = {{m{v^2}} \over r}
⇒mv2r=kr3\Rightarrow {{m{v^2}} \over r} = {k \over {{r^3}}}

⇒\Rightarrow

mv2=kr2m{v^2} = {k \over {{r^2}}}

We know kinetic energy of particle, K =

12mv2{1 \over 2}m{v^2}

=

k2r2{k \over {2{r^2}}}

As Total energy = Kinetic energy + Potential energy So Total energy =

k2r2{k \over {2{r^2}}}
−k2r2- {k \over {2{r^2}}}

= 0

Q74
A boy is rolling a 0.5 kg ball on the frictionless floor with the speed of 20 ms-1. The ball gets deflected by an obstacle on the way. After deflection it moves with 5% of its initial kinetic energy. What is the speed of the ball now?
A 14.41 ms−-1
B 19.0 ms−-1
C 4.47 ms−-1
D 1.00 ms−-1
Correct Answer
Option C
Solution
K.E.f=5% KEiK.E{._f} = 5\% \,K{E_i}
12mv2=5100×12×m×202{1 \over 2}m{v^2} = {5 \over {100}} \times {1 \over 2} \times m \times {20^2}
v2=120×202=20{v^2} = {1 \over {20}} \times {20^2} = 20
v=20=25v = \sqrt {20} = 2\sqrt 5

m/s = 4.47 m/s

Q75
A spherical ball of mass 2020 kgkg is stationary at the top of a hill of height 100100 mm. It rolls down a smooth surface to the ground, then climbs up another hill of height 3030 mm and finally rolls down to a horizontal base at a height of 2020 mm above the ground. The velocity attained by the ball is
A 2020 m/sm/s
B 4040 m/sm/s
C 1030   m/s10\sqrt {30} \,\,\,m/s
D 10  m/s10\,\,m/s
Correct Answer
Option B
Solution

Loss in potential energy == gain in kinetic energy

m×g×80=12mv2m \times g \times 80 = {1 \over 2}m{v^2}

⇒\Rightarrow

10×80=12v210 \times 80 = {1 \over 2}{v^2}

⇒\Rightarrow

v2=1600{v^2} = 1600

or

v=40 m/sv = 40\,m/s
Q76
A body of mass m starts moving from rest along x-axis so that its velocity varies as υ=as\upsilon = a\sqrt s where a is a constant and s is the distance covered by the body. The total work done by all the forces acting on the body in the first t seconds after the start of the motion is :
A 18 {1 \over 8}\, m a4 t2
B 8 m a4 t2
C 4 m a4 t2
D 14 {1 \over 4}\, m a4 t2
Correct Answer
Option A
Solution

Given,

υ\upsilon

= a

s\sqrt s

⇒\Rightarrow

   \,\,\,
dsdt=as{{ds} \over {dt}} = a\sqrt s

⇒\Rightarrow

   \,\,\,
∫0tdss=∫0za dt\int\limits_0^t {{{ds} \over {\sqrt s }}} = \int\limits_0^z {a\,dt}

⇒\Rightarrow

   \,\,\,

2

s\sqrt s

= at ⇒\Rightarrow

   \,\,\,

s =

a2t24{{{a^2}{t^2}} \over 4}

=

12.a22.t2{1 \over 2}.{{{a^2}} \over 2}.{t^2}
∴    \therefore\,\,\,\,

acceleration =

a22{{{a^2}} \over 2}
∴   \therefore\,\,\,

Force (F) = m ×\times

a22{{{a^2}} \over 2}
∴    \therefore\,\,\,\,

Work done = F. S =

ma22×a2t24{{m{a^2}} \over 2} \times {{{a^2}{t^2}} \over 4}

=

ma4t28{{m{a^4}{t^2}} \over 8}
Q77
Two particles of the same mass m are moving in circular orbits because of force, given by F(r)=−16r−r3F\left( r \right) = {{ - 16} \over r} - {r^3} The first particle is at a distance r = 1, and the second, at r = 4. The best estimate for the ratio of kinetic energies of the first and the second particle is closest to :
A 6×10−26 \times {10^{ - 2}}
B 3×10−33 \times {10^{ - 3}}
C 10−1{10^{ - 1}}
D 6×1026 \times {10^{ 2}}
Correct Answer
Option A
Solution

In circular motion the force required

∣F∣=mv2r\left| F \right| = {{m{v^2}} \over r}
∴   \therefore\,\,\,
mv2r=16r+r3{{m{v^2}} \over r} = {{16} \over r} + {r^3}

⇒\Rightarrow mv2 = 16 + r4

∴   \therefore\,\,\,

kinetic energy (K) =

12{1 \over 2}

mv2 =

12{1 \over 2}

[ 16 + r4]

∴   \therefore\,\,\,

Kinetic energy of first particle (K1) =

12{1 \over 2}

[16 + 1] Kinetic energy of second particle (K2) =

12{1 \over 2}

[16 + 44]

∴    \therefore\,\,\,\,
K1K2{{{K_1}} \over {{K_2}}}

=

16+1216+2562{{{{16 + 1} \over 2}} \over {{{16 + 256} \over 2}}}

=

17272{{17} \over {272}}

⇒\Rightarrow

   \,\,\,
K1K2=6×10−2{{{K_1}} \over {{K_2}}} = 6 \times {10^{ - 2}}
Q78
A particle of mass mm moves on a straight line with its velocity increasing with distance according to the equation v=αxv=\alpha \sqrt{x}, where α\alpha is a constant. The total work done by all the forces applied on the particle during its displacement from x=0x=0 to x=dx=\mathrm{d}, will be :
A m2α2 d\dfrac{\mathrm{m}}{2 \alpha^2 \mathrm{~d}}
B md2α2\dfrac{\mathrm{md}}{2 \alpha^2}
C mα2 d2\dfrac{\mathrm{m} \alpha^2 \mathrm{~d}}{2}
D 2 mα2 d2 \mathrm{~m} \alpha^2 \mathrm{~d}
Correct Answer
Option C
Solution

To find the total work done by all forces applied on the particle during its displacement, we can use the work-energy theorem which states that the work done by all forces on an object is equal to the change in kinetic energy of the object.

So, we first need to find the initial and final kinetic energies of the particle and then calculate the work done.

The velocity of the particle is given by

v=αxv = \alpha \sqrt{x}

, and the kinetic energy

KK

of the particle is given by

K=12mv2K = \frac{1}{2} m v^2

. We can substitute the expression for

vv

into this formula to get the kinetic energy as a function of position

xx

:

K(x)=12m(αx)2=12mα2xK(x) = \frac{1}{2} m (\alpha \sqrt{x})^2 = \frac{1}{2} m \alpha^2 x

To find the total work done from

x=0x = 0

to

x=dx = d

, we need to compute the difference in kinetic energy between these two points:

W=K(d)−K(0)W = K(d) - K(0)

At

x=dx = d

,

K(d)=12mα2dK(d) = \frac{1}{2} m \alpha^2 d

At

x=0x = 0

, since the particle starts from this position,

K(0)=12mα2(0)=0K(0) = \frac{1}{2} m \alpha^2 (0) = 0

So, the work done

WW

is simply the kinetic energy at

x=dx = d

,

W=12mα2d−0=12mα2dW = \frac{1}{2} m \alpha^2 d - 0 = \frac{1}{2} m \alpha^2 d

This matches with Option C:

mα2d2\frac{m \alpha^2 d}{2}

.

Q79
A block moving horizontally on a smooth surface with a speed of 40 m/s splits into two parts with masses in the ratio of 1 : 2. If the smaller part moves at 60 m/s in the same direction, then the fractional change in kinetic energy is :-
A 13{{1 \over 3}}
B 23{{2 \over 3}}
C 18{{1 \over 8}}
D 14{{1 \over 4}}
Correct Answer
Option C
Solution

3MV0 = 2MV2 + MV1 3V0 = 2V2 + V1 120 = 2V2 + 60 ⇒\Rightarrow V2 = 30 m/s

ΔK.E.K.E.=12MV12+122MV22−123MV02123MV02{{\Delta K.E.} \over {K.E.}} = {{{1 \over 2}MV_1^2 + {1 \over 2}2MV_2^2 - {1 \over 2}3MV_0^2} \over {{1 \over 2}3MV_0^2}}
=V12+2V22−3V023V02= {{V_1^2 + 2V_2^2 - 3V_0^2} \over {3V_0^2}}
=3600+1800−48004800=18= {{3600 + 1800 - 4800} \over {4800}} = {1 \over 8}
Q80
A body of mass m = 10–2 kg is moving in a medium and experiences a frictional force F = –kv2. Its initial speed is v0 = 10 ms–1. If, after 10 s, its energy is 18mv02{1 \over 8}mv_0^2, the value of k will be:
A 10-1 kg m-1 s-1
B 10-3 kg m-1
C 10-3 kg s-1
D 10-4 kg m-1
Correct Answer
Option D
Solution

According to the question, final kinetic energy =

18mv02{1 \over 8}mv_0^2

Let final speed of the body = Vf So final kinetic energy =

12mvf2{1 \over 2}mv_f^2

According to question,

12mvf2{1 \over 2}mv_f^2

=

18mv02{1 \over 8}mv_0^2
⇒vf=v02\Rightarrow {v_f} = {{{v_0}} \over 2}

=

102{{10} \over 2}

= 5 m/s Given that, F = –kv2 ⇒\Rightarrow

m(dvdt)m\left( {{{dv} \over {dt}}} \right)
=−kv2= - k{v^2}
⇒10−2(dvdt)=−kv2\Rightarrow {10^{ - 2}}\left( {{{dv} \over {dt}}} \right) = - k{v^2}
⇒∫105dvv2=−100k∫010dt\Rightarrow \int\limits_{10}^5 {{{dv} \over {{v^2}}}} = - 100k\int\limits_0^{10} {dt}
⇒15−110=100k×10\Rightarrow {1 \over 5} - {1 \over {10}} = 100k \times 10
⇒k=10−4kg m−1\Rightarrow k = {10^{ - 4}}kg\,{m^{ - 1}}
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