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 ?
Concept of Physics - 1 , HC VERMA , Chapter " Work and Energy"
Answers
Answered by
64
Given in the question :-
Mass of the stone m :- 200 g = 0.2 kg.
Height from the ground, h :- 150 cm = 1.5 m
speed = 3 m/s
t = 1 sec.
Now work done against gravity for lifting the stone :-
= mgh
= 0.2 × 9.8 × 1.5
= 2.94 J.
Thus Kinetic Energy of the stone = 1/2 (mv²)
= 1/2 (0.2 × 3² )
= 1.8/2
=0.90 J.
Hence total work done by demonstrator = Work done against gravity + K.E of stone
=2.94 + 0.90
= 3.84 J.
Now power by demonstrator = W/t
P = 3.84 / 1
P= 384 W.
Changing in horsepower.
=384/746 [ 1 hp = 746 W]
= 0.00514
P= 5.14 × 10⁻³ hp.
Hope it Helps.
Mass of the stone m :- 200 g = 0.2 kg.
Height from the ground, h :- 150 cm = 1.5 m
speed = 3 m/s
t = 1 sec.
Now work done against gravity for lifting the stone :-
= mgh
= 0.2 × 9.8 × 1.5
= 2.94 J.
Thus Kinetic Energy of the stone = 1/2 (mv²)
= 1/2 (0.2 × 3² )
= 1.8/2
=0.90 J.
Hence total work done by demonstrator = Work done against gravity + K.E of stone
=2.94 + 0.90
= 3.84 J.
Now power by demonstrator = W/t
P = 3.84 / 1
P= 384 W.
Changing in horsepower.
=384/746 [ 1 hp = 746 W]
= 0.00514
P= 5.14 × 10⁻³ hp.
Hope it Helps.
Answered by
16
GIVEN:-
m = 200g = 0.2kg
h = 150cm = 1.5m
v = 3m/sec
t = 1 sec
Total work done = ½ mv^2 + mgh
= (1/2) × (0.2) ×9 + (0.2) × (9.8) × (1.5)
= 3.84 J
P = W/t
= 3.84/1
= 3.84 W
1 W = 1/746 HP
3.84 W = 3.84/746 HP
= 5.14 × 10^-3 HP
therefore, work done is 3.84 J and
power in horsepower is 5.14×10^-3 HP
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