A drop of radius 2 mm falls to the ground from a height of 500 m. With decreasing acceleration it falls to half the height from the original height where it reaches the final speed and then it moves at a steady speed. What will be the work done by the force of gravity during the second half? If it completes its journey at a speed of 10 meters per second and falls to the ground, how much work will be done by the force of the barrier during this journey?
Answers
Answer:
Given:
Radius of the rain drop r=2mm=2×10
−3
m
Density of water, ρ=10
3
kgm
−3
The mass contained in a rain drop,
m=ρV
⇒
3
4
×3.141×(2×10
−3
)
3
×10
3
kg
Gravitational force experienced by the rain drop,
F=mg
⇒
3
4
×3.141×((2×10
−3
)
3
×10
3
×9.8)N
The work done by the gravity on the drop is given by:
W
1
=Fs
=
3
4
×3.14×(2×10
3
)
3
×10
3
×9.8×250=0.082J
The work done on the drop in the second half of the journey will be same as that in the first half.
∴W
2
=0.082 J
The total energy of the drop remains conserved during its motion.
Total energy at the top:
E
T
=mgh+0
=4/3×3.141×(10
−3
)
3
×10
3
×500×10
−5
=0.164J
Due to the presence of a resistive force, the drop hits the ground with a velocity of 10m/s.
∴Total energy at the ground:
E
G
=
2
1
mv
2
+0=1/2×4/3×3.141×(2×10
−3
)
3
×10
3
×9.8×10
2
=1.675×10
−3
J
So, the work done by the resistive force = E
G
−E
T
=−0.162J