Question

A tubewell pumps out 2400 kg of water per minute If water is moving with a velocity of 3 ms what is power of the pump
how much work is done if pump runsfor 10 hrs
(ii)A car weighing 1120 kg is going up an incline of 1 in 56 at rate of 20 m in 2 s find power of engine if frictional froce is 64 n
(iii)a train weighing 100 metric tonne is running on a level track with a uniform speed of 72 kmh if frictional resistance amounts to 0.5 kg per metric tonne find power of engine
(iv)a pump can throw 10 quintals of coal per hrfrom a cola mine 120 m depth calculate power of engine in watt assuming that efficiency is 80%

Answer 1

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1.

• 2400 kg  of water per minute is pumped.

• initially water is at rest. the change in energy is the work done.

• speed of water = 3 meters/second = v

• So 2400 kg of water comes out in 1 min. 

• Hence 40 kg per second.

• let t = time. 

• so  mass/time = m/t = 40 kg/sec.

•    The kinetic energy = work done = 1/2 m v²

•    Power = work done / time = 1/2 (m / t) v2    = 1/2 * 40 kg/sec * 3² m²/sec²   = 180 kg-m sec²-m = 180 W

sec²-m = 180 Wwork done for 10 hrs = 180 * 10 * 3600 Joules.

2.

• m = 1120 kg

•      let the  angle of slope = Ф   

•   =>  tan Ф = 1/56

•      As the tan Ф is very small, angle Ф also very small.

•   So Sin Ф is almost equal to tan Ф ie., 1/5sec.

•      The car travels 20 m along the slope in 2 sec.

• => velocity = 10 m/s along the inclined plane.

    

•   Force exerted by the engine along the slope

•      = m g Sin Ф + Friction force.

•      = 1120 kg * 10 m/sec² * 1/56 + 60 N.

•      = 260 N.

Power by the engine = force * velocity = 2600 N-m/s or Watts.

3.

• As the truck is moving with uniform velocity and no acceleration, the force exerted by the engine of truck is equal to the frictional force.

•    frictional force = 0.5 kg force / metric tonne  * 100 metric tonnes.

•                          = 50 kg force = 50 * g  Newtons.

•                        = 500 Newtons.

•    Power = force * velocity in the same direction

• = 500 N * 72 * 5/18 m/sec

      

= 10,000 W = 10 kW

4.

• height = h = 120m,  

• mass = m = 10

• quintals = 1000 kg,

• time = t = 1 hr = 3600 sec.

•    change in the potential energy of coal = m g h

•      power at 100% efficiency = energy / time =  m g h / t = (m/t) g h

•             = 1000 kg / 3600sec * 10 m/sec² * 120 m

•            = 333.333 Watts

•      power at 80% efficiency = power / efficiency

• = 333.333 / (80/100)

           = 416.66 Watts

•      if you calculate the power  with  g = 9.8 m/sec² instead of 10 m/sec² .

• then you will get the power as  408.33  Watts.

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Answer 2

1.

• 2400 kg  of water per minute is pumped.

• initially water is at rest. the change in energy is the work done.

• speed of water = 3 meters/second = v

• So 2400 kg of water comes out in 1 min. 

• Hence 40 kg per second.

• let t = time. 

• so  mass/time = m/t = 40 kg/sec.

•    The kinetic energy = work done = 1/2 m v²

•    Power = work done / time = 1/2 (m / t) v2    = 1/2 * 40 kg/sec * 3² m²/sec²   = 180 kg-m sec²-m = 180 W

sec²-m = 180 Wwork done for 10 hrs = 180 * 10 * 3600 Joules.

2.

• m = 1120 kg

•      let the  angle of slope = Ф   

•   =>  tan Ф = 1/56

•      As the tan Ф is very small, angle Ф also very small.

•   So Sin Ф is almost equal to tan Ф ie., 1/5sec.

•      The car travels 20 m along the slope in 2 sec.

• => velocity = 10 m/s along the inclined plane.

    

•   Force exerted by the engine along the slope

•      = m g Sin Ф + Friction force.

•      = 1120 kg * 10 m/sec² * 1/56 + 60 N.

•      = 260 N.

Power by the engine = force * velocity = 2600 N-m/s or Watts.

3.

• As the truck is moving with uniform velocity and no acceleration, the force exerted by the engine of truck is equal to the frictional force.

•    frictional force = 0.5 kg force / metric tonne  * 100 metric tonnes.

•                          = 50 kg force = 50 * g  Newtons.

•                        = 500 Newtons.

•    Power = force * velocity in the same direction

• = 500 N * 72 * 5/18 m/sec

      

= 10,000 W = 10 kW

4.

• height = h = 120m,  

• mass = m = 10

• quintals = 1000 kg,

• time = t = 1 hr = 3600 sec.

•    change in the potential energy of coal = m g h

•      power at 100% efficiency = energy / time =  m g h / t = (m/t) g h

•             = 1000 kg / 3600sec * 10 m/sec² * 120 m

•            = 333.333 Watts

•      power at 80% efficiency = power / efficiency

= 333.333 / (80/100)

           = 416.66 Watts

•      if you calculate the power  with  g = 9.8 m/sec² instead of 10 m/sec² .

• then you will get the power as  408.33  Watts.