Solutions to Exercises on Work and Energy HC Verma's Concepts of Physics Part 1 (24-28)

 Q#24

A water pump lifts water from a level 10 m below the ground. Water is pumped at a rate of 30 kg/minute with negligible velocity. Calculate the minimum horsepower the engine should have to do this.            

Answer:
Since pumped water has negligible velocity, its K.E. is zero. The pump has to do work in increasing the P.E. of water to  10 m above.

Water pumped per second = 30 kg/60 s = 0.5 kg/s
Weight of this water = 0.5 x 9.8 N = 4.9 N

So the pump has to apply 4.9 N force against the gravity to lift it. Hence work done by the pump = 4.9 N x 10 m = 49 J/s = 49 W

So, minimum power of the engine = 49 W = 49/746 hp = 0.066 hp = 6.6x  10^-2 hp

Q#25
An unruly demonstrator lifts a stone of mass 200 g from the ground and throws it at his opponent. At the time of projection, the stone is 150 cm above the ground and has a speed of 3.00 m/s. Calculate the work done by the demonstrator during the process. If it takes 1 second for the demonstrator to lift the stone and throw, what horsepower does he use?            

Answer:
Mass of the stone m = 200 g = 0.20 kg
Height from the ground h = 150 cm = 1.50 m
So work done against gravity in lifting the stone = mgh = 0.20 x 9.8 x 1.50 = 2.94 J
Speed given to the stone v = 3.00 m/s

K.E. = ½mv² =0.50 x 0.20 x 3² = 0.90 J

So work done in increasing the K.E. of the stone = 0.90 J

Hence Total work done by the demonstrator= 2.94 J + 0.90 J = 3.84 J

Time taken in this process = 1 s

So power used by the demonstrator

= 3.84 J/1 s = 3.84 W = 3.84/746 hp = 5.15 x 10^-3 hp      

Q#26
In a factory, it is desired to lift 2000 kg of metal through a distance of 12 m in 1 minute. Find the minimum horsepower of the engine to be used.A            

Answer:
Minimum work is done by the engine when the metal is lifted slowly.
Mass of the metal = 2000 kg, Distance of the lift = 12 m.

So work done against the gravity = mgh = 2000 x 9.8 x 12 J = 235.2 kJ
Time taken = 1 min = 60 s 

Minimum horsepower required

= 235.2 kJ/60 W = 3920 W
= 3920/746 hp = 5.25 hp

Q#27
A scooter company gives the following specifications about its product
Weight of the scooter – 95 kg
Maximum speed – 60 km/h
Maximum engine power – 3.5 hp
Pickup time to get the maximum speed – 5 s
Check the validity of these specifications.            

Answer:
The engine makes the scooter to change the speed from zero to v = 60 km/h = 60000/3600 m/s = 16.67 m/s

K.E. of the scooter at rest = 0
K.E. of the scooter at maximum speed = ½mv² = 0.5 x 95 x (16.67)² = 13194.44 J
Work done by the engine = Change in K.E. = 13194.44 J

Time taken = 5 s
Required Power of the engine = 12194.44/5 = 2638.88 W = 2638.88/746 hp = 3.54 hp
But the quoted power = 3.5 hp

So, with 3.5 hp power of the engine, the scooter cannot achieve exactly 60 km/h speed in 5 s. It is somewhat overclaimed.

Q#28

A block of mass 30.0 kg is being brought down by a chain. If the block acquires a speed of 40.0 cm/s in dropping down 2.00 m, find the work done by the chain during the process.            

Answer:
Let force applied by the chain = T Newton (Upward)
Weight of the block = 30.0 kg x 9.8 m/s² = 294 N (Downward)
Final speed v = 40.0 cm/s = 0.40 m/s

Change in K.E. =½ x 30 x (0.40)² = 2.4 J

So, work done by all the forces on the block = 2.4 J

Forces on the block are force by chain and force of gravity i.e. Net force (downward) = (294 – T)  N
Work done by this force in dropping the block by 2.00 m

(294 – T) x 2.00 = 2.4
294 – T = 1.2
T = 294 – 1.2 = 292.8 N

The distance travelled by the block is opposite to the direction of the force applied by the chain = –2.00 m

So, work done by the chain during the process = 292.8 N x (–2.00 m) ≈ –586 J  

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