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Byju's Answer
Standard X
Physics
Multiple Movable Pulley
An elevator w...
Question
An elevator which can carry a maximum load of 1800kg (elevator+passengers)is moving up with a constant speed of 2m/s. The frictional force opposing the motion is 4000N. Determine the minimum power delivered by the motor to the elevator in watt and horsepower .
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Solution
Dear Student,
1
.
w
e
i
g
h
t
i
n
t
h
e
e
l
e
v
a
t
o
r
,
w
=
m
g
w
=
1800
×
9
.
8
=
17640
N
t
o
t
a
l
f
o
r
c
e
a
c
t
i
n
g
d
o
w
n
w
a
r
d
=
w
e
i
g
h
t
+
f
r
i
c
t
i
o
n
a
l
f
o
r
c
e
F
=
w
+
f
F
=
17640
+
4000
F
=
21640
N
p
o
w
e
r
d
e
l
i
v
e
r
e
d
b
y
m
o
t
o
r
=
f
o
r
c
e
×
v
e
l
o
c
i
t
y
P
=
F
×
v
P
=
21640
×
2
P
=
43280
W
a
t
t
s
We know that 1 horse power = 746 Watt
so Power = 43280/746 = 58.01 Hp
Regards
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