Q 1 :

A body is moving unidirectionally under the influence of a constant power source. Its displacement in time t is proportional to                    [2024]

  • t

     

  • t2

     

  • t2/3

     

  • t3/2

     

(4)     

           dWdt=P⇒dW=Pdt

            F·ds=Pdt

           mdvdsdt=Pdt

          ∫0vvdv=Pm∫0tdt⇒v22=Ptm

           Hence,  v∝t⇒dsdt=Ct, C is a constant

           ∫0sds=C∫0tt1/2dt

          s=C23t3/2⇒s∝t3/2

       ∴    displacement of proportional to (t)3/2

 



Q 2 :

A body of mass 2 kg begins to move under the action of a time dependent force given by F→=(6ti^+6t2j^)N. The power developed by the force at the time t is given by           [2024]

  • (6t4+9t5)W

     

  • (3t5+6t5)W

     

  • (9t5+6t3)W

     

  • (9t3+6t5)W

     

(4)     

          F→=(6ti^+6t2j^)N

          F→=ma→=(6ti^+6t2j^)

          a→=F→m=(3ti^+3t2j^)

          v→=∫0ta→dt=3t22i^+t3j^

         P=F→·v→=(9t3+6t5)W

 



Q 3 :

A body of mass 4 kg is placed on a plane at a point P having coordinate (3, 4) m. Under the action of force F→=(2i^+3j^)N, it moves to a new point Q having coordinates (6, 10) m in 4 sec. The average power and instantaneous power at the end of 4 sec are in the ratio of :          [2025]

  • 13 : 6

     

  • 6 : 13

     

  • 1 : 2

     

  • 4 : 13

     

(2)

F→=(2i^+3j^)N; displacement S→=(3i^+6j^)m

Acceleration, a→=F→m=(2i^+3j^4)m/s2

                              S→=u→t+12a→t2

⇒ 3i^+6j^=4u→+12(2i^+3j^4)·16=4u→+4i^+6j^

or u→=14(3i^+6j^–4i^–6j^) ⇒ u→=–14i^

Average Power <P>=F→·S→t

                                                =(2i^+3j^)·(3i^+6j^)4watt=6watt

                                  v→ at t=4sec = (12i^+34j^)×4=(2i^+3j^)

                                      Pins=F→·v→=(2i^+3j^)(2i^+3j^)=13 W

⇒ <P>Pins=613

 



Q 4 :

A sand dropper drops sand of mass m(t) on a conveyer belt at a rate proportional to the square root of speed (v) of the belt, i.e., dmdt∝v. If P is the power delivered to run the belt at constant speed then which of the following relationship is true?          [2025]

  • P2∝v3

     

  • P∝v

     

  • P∝v

     

  • P2∝v5

     

(4)

Power = F→·v→ and F=dpdt=Rate of change of linear momentum

              F=v·dmdt=Kv3/2, K is constant

Power (P)=(Kv)3/2·(v)=Kv5/2

So, P2∝v5



Q 5 :

An object of mass 1000 g experiences a time dependent force F→=(2ti^+3t2j^)N. The power generated by the force at time t is:        [2025]

  • (2t3+3t3)W

     

  • (2t2+18t3)W

     

  • (3t3+5t5)W

     

  • (2t3+3t5)W

     

(4)

m = 1000 gram = 1 kg

F→=(2ti^+3t2j^)

dvdt=Fm=(2ti^+3t2j^)

⇒ ∫0vdv=∫0t(2ti^+3t2j^)dt

⇒ v→=t2i^+t3j^

Power, P=F→·v→

P=(2ti^+3t2j^)·(t2i^+t3j^)

P=(2t3+3t5)W



Q 6 :

The ratio of powers of two motors is 3xx+1, that are capable of raising 300 kg water in 5 minutes and 50 kg water in 2 minutes respectively from a well of 100 m deep. The value of x will be                   [2023]
 

  • 2

     

  • 4

     

  • 2.4

     

  • 16

     

(4)

Average Power=total work donetotal time=mght

P1P2=m1ght1m2ght2=m1t1t2m2

P1P2=300×25×50=125=3xx+1

⇒x=16



Q 7 :

A body of mass 1 kg begins to move under the action of a time dependent force F→=(ti^+3t2j^)N. where i^ and j^ are the unit vectors along x and y axis. The power developed by the above force, at the time t=2 s will be _______ W.          [2023]



(100)

F→=ti^+3t2j^

mdv→dt=ti^+3t2j^

m=1 kg, ∫0vdv=∫0ttdti^+∫0t3t2dtj^

v→=t22i^+t3j^

Power=F→·v→=t32+3t5

At t=2, Power=82+3×32=100



Q 8 :

A body of mass 2 kg is initially at rest. It starts moving unidirectionally under the influence of a source of constant power P. Its displacement in 4 sec is 13α2P m. The value of α will be ________ .               [2023]



(4)

12mv2=Pt

v=2Ptm

dxdt=2Ptm

x=2Pm23[t3/2]04

x=16P3=13×16P

⇒α=4



Q 9 :

If the maximum load carried by an elevator is 1400 kg (600 kg – passenger + 800 kg – elevator), which is moving up with a uniform speed of 3 ms-1 and the frictional force acting on it is 2000 N, then the maximum power used by the motor is _______ kW (g=10 m/s2)           [2023]



(48)

Pmax=Fmax×v

Fmax=1400 g+friction

          =14000+2000=16000

Pmax=16000×3=48000 W=48 kW



Q 10 :

A block of mass 5 kg starting from rest pulled up on a smooth inclined plane making an angle of 30° with the horizontal with an effective acceleration of 1 ms-2. The power delivered by the pulling force at t=10 s from the start is ________ W. [Use g=10 ms-2]  (calculate the nearest integer value)            [2023]



(300)

F-5sin30°=5a

⇒F=5+25=30 N

       v10=u+at

⇒v10=0+1(10)=10 m/s

      P10=Fv=300 W



Q 11 :

A body of mass 2 kg begins to move under the action of a time dependent force given by F→=(6ti^+6t2j^)N. The power developed by the force at the time t is given by:

  • (6t4+9t5) W

     

  • (3t3+6t5) W

     

  • (9t5+6t3) W

     

  • (9t3+6t5) W

     

(4)

F→=(6ti^+6t2j^) N

F→=ma→=(6ti^+6t2j^)

a→=F→m=(3ti^+3t2j^)

v→=∫0ta→ dt=3t22i^+t3j^

P=F→·v→=(9t3+6t5) W



Q 12 :

A particle (bob) of mass 1 kg is performing vertical circular motion. Then:

  • Power delivered by tension at point (4) is negative.

     

  • Power delivered by tension at point (2) is positive.

     

  • Power delivered by gravitational force at point (3) is zero.

     

  • Power delivered by gravitational force at point (4) is negative.

     

(4)

P=F→·v→,    Tension is always perpendicular to velocity



Q 13 :

A body of mass 2 kg begins to move under the influence of time dependent force F→=(2ti^+6t2j^) N, where i^ and j^ are unit vectors along x and y-axis respectively. The power produced by the force at t=2 s is _______ W.                     [2026]



(200)

Given: mass of body=2 kg

F→=(2ti^+6t2j^) N

acceleration  a→=F→m=2ti^+6t2j^2=(ti^+3t2j^) ms-2

v→=∫02a→·dt=[t22i^+3t33j^]02=(2i^+8j^) m/s

F→t=2s=(4i^+24j^) N

P=F→·v→=8+192=200 W