Asked by sam
A person weighing 0.6 kN rides in an elevator that has a downward acceleration of 2.3 m/s^2.
The acceleration of gravity is 9.8 m/s^2.
What is the magnitude of the force of the elevator floor on the person?
Answer in units of kN
The acceleration of gravity is 9.8 m/s^2.
What is the magnitude of the force of the elevator floor on the person?
Answer in units of kN
Answers
Answered by
Damon
m = 600/g = 61.2 kg
force up on object = F
force down on object = m g = 600 N
net force up = F-600
F = m a
F-600 = 61.2 (-2.3)
F = 600 - 141 = 459 N = .459 kN
force up on object = F
force down on object = m g = 600 N
net force up = F-600
F = m a
F-600 = 61.2 (-2.3)
F = 600 - 141 = 459 N = .459 kN
Answered by
Salmon
Since you know the Fg (which is 0.6kN * 1000 = 600N), you can use that to find the mass of the object:
mg = Fg
mg = 600
m(9.8) = 600
m = 61.2
Now, you use the sum of forces formula, keeping in mind that you have 2 forces in the vertical direction: Ft and Fg. MAKE DOWN THE POSITIVE DIRECTION, THIS WILL HELP YOU LATER ON
F = ma
Fg - Ft = (61.2)(2.3) --> (acceleration was given, and Fg is positive b/c we made down the positive direction)
600 - Ft = (61.2)(2.3)
Ft = 459 N ==> 0.459 kN
mg = Fg
mg = 600
m(9.8) = 600
m = 61.2
Now, you use the sum of forces formula, keeping in mind that you have 2 forces in the vertical direction: Ft and Fg. MAKE DOWN THE POSITIVE DIRECTION, THIS WILL HELP YOU LATER ON
F = ma
Fg - Ft = (61.2)(2.3) --> (acceleration was given, and Fg is positive b/c we made down the positive direction)
600 - Ft = (61.2)(2.3)
Ft = 459 N ==> 0.459 kN
Answered by
John
thanks
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