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
work 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., 1/56.
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.
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.