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JEE Main Physics Class 11 Work, Energy and Power Part 2 Questions
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© examsnet.com
Question : 1
Total: 100
If the maximum load carried by an elevator is
1400
kg
(
600
kg
−
Passengers
+
800
kg
- elevator), which is moving up with a uniform speed of
3
m
s
−
1
and the frictional force acting on it is
2000
N
, then the maximum power used by the motor is _______
kW
(
g
=
10
m
∕
s
2
)
[10-Apr-2023 shift 2]
Your Answer:
Validate
Solution:
Tension in the string
⇒
16000
N
Maximum power
=
(
F
)
(
V
)
=
16000
×
3
=
48000
=
48
kW
Ans. 48
© examsnet.com
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