A machine lifts an object of weight 200.N at constant speed of 2m⋅s−1 through a vertical distance of 4m. The efficiency of the machine is 25 %. What is the input power of the machine? Give your answer in kW, correct to two significant figures. Remember to include a unit with your answer.
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Answer:
The work going into mechanical energy is W= KE + PE. At the bottom of the stairs, we take both KE and PEg as initially zero; thus,
W
=
KE
f
+
PE
g
=
1
2
m
v
2
f
+
m
g
h
, where h is the vertical height of the stairs. Because all terms are given, we can calculate W and then divide it by time to get power.
Solution
Substituting the expression for W into the definition of power given in the previous equation,
P
=
W
t
yields
P
=
W
t
=
1
2
m
v
2
f
+
m
g
h
t
Entering known values yields
P
=
0.5
(
60.0
kg
)
(
2.00
m/s
)
2
+
(
60.0
kg
)
(
9.80
m/s
2
)
(
3.00
m
)
3.50
s
=
120
J
+
1764
J
3.50
s
=
538
W
Discussion
The woman does 1764 J of work to move up the stairs compared with only 120 J to increase her kinetic energy; thus, most of her power output is required for climbing rather than accelerating.
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