Tag: energy and its forms

Quizzes (9) Questions (123)

Questions Related to energy and its forms

Multiple choice power work and power work, energy and power physics energy and its forms

A force applied by the engine of a train of mass 2.05 x $10^{6} kg$ changes its velocity from 5 $ms^{-1}$ to 25  $ms^{-1}$ in 5 minutes. The power of the engine is then

  1. 1.025 MW

  2. 2.05 MW

  3. 5 MW

  4. 6 MW
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given that,

Mass of train, $m=2.05\times {{10}^{6}}\,Kg$

Time, $t=5\,\min utes=300\,s$

$ v=25\,m/s $

$ u=5\,m/s $

Acceleration,

$ a=\dfrac{v-u}{t} $

$ a=\dfrac{25-5}{300} $

$ a=\dfrac{2}{30}=\dfrac{1}{15}\,m/{{s}^{2}} $

Using equation of motion,

$ {{v}^{2}}-{{u}^{2}}=2as $

$ {{(25)}^{2}}-{{(5)}^{2}}=2\times \dfrac{1}{15}\times s $

$ s=4500\,m $

Power,

$ P=\dfrac{work\,\,done}{time} $

$ P=\dfrac{F\times d}{t} $

$ P=\dfrac{m\times a\times s}{t} $

$ P=\dfrac{2.05\times {{10}^{6}}\times 1\times 4500}{15\times 300} $

$ P=2.05\times {{10}^{6}}\,W $

$ P=2.05\,MW $

Multiple choice power work and power work, energy and power physics energy and its forms

A motor lifts $100 kg$ of water in $2 min$ from a well of $60m$ depth then the electric power of the motor is$(Taken g=10 m/s^2)$

  1. $1000 W$
  2. $750 W$
  3. $1200 W$
  4. $500W$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A motor lifts $=100kg$ of water

Time $=2min=2\times60=120s$
Depth$=60m$ depth then,
electric power of the motor$=?$
Taking $=10m/s^2$
$P=Power=\cfrac{mgh}{t}\ \quad=\cfrac{100\times10\times60}{120}\ \quad=500W$

Multiple choice power work and power work, energy and power physics energy and its forms

A force $'F'$ accelerates a block of mass $'m'$ along a straight line to velocity $'v'$ from rest and displace it through a distance $'s'$. What is the average power developed?

  1. $\dfrac {v^{2}}{2F}$
  2. $Fv$
  3. $\dfrac {mv^{2}}{2s}$
  4. $\dfrac {Fv}{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For constant acceleration starting from rest, v_avg = v / 2. Average Power = F * v_avg = F * (v / 2).

Multiple choice power work and power work, energy and power physics energy and its forms

A pump ejects $12000kg$ of water at speed of $4m/s$ in $40$ second. Find the average rate at which the pump is working

  1. $0.24KW$
  2. $2.4KW$
  3. $24KW$
  4. $24W$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Force = change in momentum / time = (m * v) / t = (12000 * 4) / 40 = 1200 N. Power = Force * velocity = 1200 * 4 = 4800 W = 4.8 kW. The provided answer 2.4 kW suggests the calculation might be (1/2) * m * v^2 / t = (0.5 * 12000 * 16) / 40 = 2400 W = 2.4 kW, which is the rate of kinetic energy delivery.

Multiple choice power work and power work, energy and power physics energy and its forms

If the average power radiated by the star is $10 ^ { 16 } \mathrm { W }$ , the deuteron supply of the star is exhausted in a time of the order of 

  1. $10 ^ { 6 }$ seccond
  2. $10 ^ { 14 }$ second
  3. $10 ^ { 12 }$ second
  4. $10 ^ { 16 }$ second
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

This is an order-of-magnitude estimation problem. Total energy available in a star's fuel (deuteron) is typically around 10^28 J. Time = Energy / Power = 10^28 / 10^16 = 10^12 seconds.

Multiple choice power work and power work, energy and power physics energy and its forms

A boat moving with constant speed v in still waters experiences a total frictional force F. The power developed by the boat is

  1. $\frac{1}{2}Fv$
  2. $Fv$
  3. $\frac{1}{2}Fv^2$
  4. $Fv^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For an object moving at constant velocity, the driving force must equal the frictional force. Power = Force * velocity = F * v.

Multiple choice power work and power work, energy and power physics energy and its forms

A body of mass m is projected at an angle $\displaystyle \theta $ with the horizontal with an initial velocity $\displaystyle v _{0}.$ The average power of gravitational force over the whole time of flight is

  1. $\displaystyle mg\cos \theta $
  2. $\displaystyle \frac{1}{2}mg\sqrt{u\cos \theta }$
  3. $\displaystyle \frac{1}{2}mgu\sin \theta $

  4. zero
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The projected body again comes back down. Hence the net displacement in the vertical direction will be 0.
$\therefore W=mgh=mg\times 0=0$