A 7300 N elevator is to be given an acceleration of 0.150g by connecting it to a cable of negligible weight wrapped around a tur
1 answer:
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The mechanical energy at top =Mechanical energy at bottom
- Mass=m=54kg
- Height=h=51m
- Acceleration due to gravity=g=10m/s^2
- Velocity=v=2.6m/s
- Final energy at bottom=The kinetic energy
Answer:
a)11.25 J
b)Number of revolution = 1
Explanation:
Given that
Radius ,r= 0.8 m
m= 0.3 kg
Initial speed ,u= 10 m/s
final speed ,v= 5 m/s
a)
Initial energy
KEi= 15 J
Final kinetic energy
KEf=3.75 J
The energy transformed from mechanical to internal = 15 - 3.75 J = 11.25 J
b)
The minimum value to complete the circular arc
Now by putting the values
V= 2.82 m/s
So kinetic energy KE
KE=1.19 J
ΔKE= KEi - KE
ΔKE= 15- 1.19 J
ΔKE=13.80 J
The minimum energy required to complete 2 revolutions = 2 x 11.25 J
= 22.5 J
Here 22.5 J is greater than 13.8 J.So the particle will complete only one revolution.
Number of revolution = 1