PHYSICS – WORK, POWER & ENERGY

OBJECTIVE  TYPE QUESTION

Question – 1.

A force of (x² + 2) N acts upon a particle to displace it from x = 3 m to x = 6 m. The work done is equal to :

(A) 63 J                                 (B) 69 J                                  (C) 72 J                                  (D) 81 J

Solution : –

We have,             work done, W   =             ₃∫⁶(x² + 2)dx

=             [x³/3 + 2x]₃⁶

=             [216/3 + 12] – [27/3 + 6]

=             84 – 15

=             69 J         (B)          [Ans.]

Question – 2.

The potential energy, U = ax³ – 2bx². The force constant would be :

(A) 4b                   (B) 3b                    (C) 2b                    (D) b

Solution : –

We have,             U             =             ax³ – 2bx²

And        Force constant, K             =             d²U/dx²

Now,                                    dU/dx   =             3ax² – 4bx

Or,                                          x              =             4b/3a

And                        d²U/dx² =             6ax – 4b

At           x              =             4b/3a,                   K             =             d²U/dx²

=             6a(4b/3a) – 4b

=             2b           (C)          [Ans.]

Question – 3.

Due to change in velocity the kinetic energy of a body is increased by 300%. The momentum is increased by :

(A) 100%                             (B) 200%                              (C) 300%                              (D) 400%.

Solution : –

We have, if initial K.E.     =             100 x ,   final K.E.              =             100 x + 300 x       =             400 x

Therefore,                          P₁²/P₂² =             2mk₁/_2mk2

=             100 x/400 x

=             1/4

Or,                                          P₂/P₁ =             2/1

Or,                                          P₂ =             2P₁

Therefore,          Increase in momentum                 =             {(P₂ – P₁)/P₁}×100 %

=             {(2P₁ – P₁)/P₁}×100 %

=             100 %    (A)          [Ans.]

Question – 4.

A particle rotates in a circular orbit of radius R with centripetal force, F = β/R². The mass of the particle being m, the total kinetic energy of the particle is :

(A) β/2R                              (B) 2R/β                               (C) – β/2R                            (D) – 2R/β.

Solution : –

We have,  Centripetal force,       F             =             β/R².

Potential Energy,U          =             _∞∫^RF.dR

=             β_∞∫^RdR/R²

=          β[– 1/R]∞R

=             – β/R

Kenetic Energy                  =             1/2mv²

=             (R/2)[(mv²)/R]

=             (R/2)(β/R²)

=             β/2R

Therefore, total kinetic energy =             K.E.        +             P.E.

=             β/2R – β/R

=             – β/2R                   (C)          [Ans.]

Question – 7.

U denotes potential energy, which is a function and depends on r as :

U             =             a/r¹⁰ – b/r⁵

Where r is the inter-atomic distance and a and b positive constant. The equilibrium distance between two atoms is :

(A) (b/2a)^1/10 (B) (b/2a)^1/5 (C) (2a/b)^1/10 (D) (2a/b)^1/5.

Solution : –

We have,             U             =             a/r¹⁰ – b/r⁵

dU/dr                    =             – 10a/r¹¹ + 5b/r⁶

For equilibrium, dU/dr   =             0

Therefore,          – 10a/r¹¹ + 5b/r⁶=             0

Or,                                          2a/r⁵ =             b

Or,                                          r              =             (2a/b)^1/5 (D)          [Ans.]

Question – 8.

A force F = Kx² acts on a particle at an angle of 60˚ with x-axis.  The work done in displacing the particle from distances x₁ to x₂ would be :

(A) K(x₂ – x₁ )             (B) K/6(x₂³ – x₁³)                               (C) K/6(x₂² – x₁²)               (D)          K(x₂³ – x₁³)

Solution : –

We have,             dW         =             F^→.dx^→

=             Fdx cos θ

=             Fdx cos 60˚

=             1/2Fdx

=             1/2Kx²

Therefore,                          W            =             K/2 _x1∫^x2 x2dx

=             K/2[x³/3]_x1^x2

=             K/6(x₂³ – x₁³)      (B)          [Ans.]

Question – 9.

A mass M is attached to an elastic wire of length L. The wire is stretched to a length of l. The mechanical energy stored in the wire would be :

(A)Mgl (B) MgL                                                (C) MgL/2                            (D) Mgl/2.

Solution : –

We have,             The mechanical energy                                 =             Work done

=             1/2 × stress × strain × volume

=             1/2 × (Mg/A)× (l/L) × (AL)

=             1/2 Mgl (C)          [Ans.]

Question – 10.

A body falls from a height of h and rebounds from hard surface such that 19% of the energy is lost in the impact. The coefficient of restitution would be :

(A)0.9                                    (B) 0.3                                   (C) 0.81                                 (D) 0.19

Solution : –

We have,             e             =             √(h₂/h₁)

And                                        mgh₂ =             (mgh₁) ×(81/100)

Or,                                          h₂ =             (81/100) × h₁

Therefore,                          e             =             √[{(81/100) × h₁}/h₁]

=             9/10

=             0.9          (A)                          [Ans.]

Question – 11.

If momentum of a particle is numerically equal to its kinetic energy then velocity of the particle would be :

(A)4 m/s                              (B) 3 m/s                              (C) 2.5 m/s                          (D) 2 m/s

Solution : –

We have,             Momentum       =             mv,        K.E.        =             1/2 mv²

As per question,               1/2 mv² =             mv

Or,                                                                          v              =             2 m/s     (D)          [Ans.]

Question – 12.

A weightless spring is compressed by a distance of α. The potential energy would be proportional to :

(A)α² (B) α                                      (C) α⁰ (D) α^–1

Solution : –

We have, in the case of a spring, F           =             kx,

where, k = spring constant and x = displacement.

Therefore,                          P.E.        =             ₀∫^αF.dx

=             ₀∫^αkxdx

=             (kx²)/2

=             k/2 × (α²)

Or,                                          P.E. is proportional to α2              (A)          [Ans.]

Question – 13.

A spring contains energy of 4 J when stretched by 2 mm. The energy contained by the same spring when stretched by 10 mm would be :

(A)10 J                                  (B) 50 J                                  (C) 100 J                               (D) none of these.

Solution : –

We have,             E₁ =             4 J,          E₂ =             ?

Using                     Energy                  =             1/2 kx²

Therefore,                          E₁ =             1/2 kx₁²

And                                        E₂ =             1/2 kx₂²

Therefore,                          E₂/E₁ =             x₂²/x₁²

=             (x₂/x₁)²

=             (10/2)²

=             25

Or,                                          E₂ =             25 E₁

But we have,     E₂ =             25 × 4

=             100 J      (C)          [Ans.]

Question – 14.

A motor car of 100 H.P. moves with a uniform speed of 72 km/h The forward thrust applied on the car by the engine would be :

(A)3730 N                            (B) 3450 N                           (C) 2260 N                           (D) 1234 N.

Solution : –

We have,             Forward thrust                  =             P/v

=             (100×746×60×60)/(72×1000)

=             3730 N (A)           [Ans.]

Question – 15.

Sand particles are falling vertically at the rate of 2 kg/s to a conveyor belt which is moving horizontally with a velocity of 0.2 m/s. The extra force required to keep the belt moving would be :

(A)0.2 N                               (B) 0.3 N                               (C) 0.4 N                               (D) 0.5 N.

Solution : –

We have,             Extra Force, F     =             d/dt(mv)

=             v.dm/dt

=             0.2 × 2

=             0.4 N      (C)          [Ans.]

Question – 16.

The K.E. of a light and heavy objects are same, the object having more momentum would be :

(A)light object                  (B) heavy object               (C) both the objects       (D) none of these.

Solution : –

We have,             1/2 m₁v₁² =             1/2 m₂v₂²

Or,                                                          v₁/v₂ =             √(m₂/m₁)

Therefore,                                          P₁/P₂ =             (m₁v₁)/(m₂v₂)

=             (m₁/m₂)√(m₂/m₁)

=             √(m₁/m₂)

When                    m₁ > m₂;               P₁ > P₂

Therefore heavier object has more momentum.                              (B)          [Ans.]

Question – 17.

A metal ball of mass 2 kg is moving with a velocity of 36 km/h and collides with a stationary ball of mass 3 kg. After collision, the moves as a single mass, the loss in kinetic energy would be :

(A)120 J                                (B) 100 J                               (C) 80 J                                  (D) 60 J.

Solution : –

We have, speed of metal ball, v₁ =             36 km/h

=             10 m/s

Using law of conservation of momentum ,           m₁v₁ + m₂v₂ =             (m₁ + m₂)v

Or,                                                          v              =             (m₁v₁ + m₂v₂ )/(m₁ + m₂)

=             {(2×10) + (3×0)}/(2 + 3)

=             4 m/s

Therefore, loss in K.E.    =             [1/2 m₁v₁² + 1/2 m₂v₂² ] – 1/2 (m₁ + m₂)v²

=             [1/2 ×2×(10)² + 1/2 ×3×(0)²] – 1/2×(2 + 3)×(4)²

=             100 – 40

=             60 J         (D)          [Ans.]

Question – 18.

A child is sitting on a swing and his minimum and maximum heights from the ground are respectively 0.75 m and 2 m. His maximum speed would be :

(A)5 m/s                              (B) 8 m/s                              (C) 12 m/s                           (D) 15 m/s

Solution : –

We have,             Maximum K.E.                   =             Drop in P.E.

Or,                                                          1/2 mv² =             mg(2 – 0.75)

Or,                                                                          v              =             √{2g(1.25)}

=             5 m/s     (A)          [Ans.]

Question – 19.

A ball of mass 2 kg and another of mass 4 kg are dropped together from a 60 m tower. After a fall of 30 m each towards the earth, their respective K.E. will be in the ratio of :

(A)1 : 4                                  (B) 2 : 3                                 (C) 1 : 2                                 (D) 1 : 3

Solution : –

We have, as initial velocity is zero.            v² =             2gs

Ratio of kinetic energies                               =             KE₁/KE₂

=             {(1/2)m₁v₁²}/{(1/2)m₂v₂²}

=             {(1/2)m₁(2gs)}/{(1/2)m₂(2gs)}

=             m₁/m₂

=             2/4

=             1 : 2        (C)          [Ans.]

Question – 20.

A particle of mass m₁ and another of mass m₂ are moving respectively with velocities v₁ and v₂. They have same momentum but their different K.E. are E₁ and E₂ respectively. If m₁ > m₂ then :

(A)E₁ = E₂ (B) E₁ < E₂ (C) E₁ > E₂ (D) none of these.

Solution : –

We have,             K.E.        =             P²/2m

Therefore,                          E₁/E₂ =             (P₁²/2m₁)/(P₂²/2m₂)

=             m₂/m₁

Now,                     m₁ > m₂ =>           E₁ < E₂ (B)          [Ans.]

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