A physics student who weighs 516.0 N stands on a bathroom scale in an elevator t
ID: 1329075 • Letter: A
Question
A physics student who weighs 516.0 N stands on a bathroom scale in an elevator that is supported by an elevator cable; the mass of the elevator, including the student inside, is 868.0 kg . While the student is standing on the scale, the elevator makes an upward trip, followed by a downward trip.
A)Find the magnitude of the acceleration of the elevator at the instant of time when the scale reads 449.0 N
B)What is the tension in the elevator cable in Part (A)?
When the Scale Reads More than the Weight of the Student
At some time during her elevator ride, the scale reads 686.0 N .->>
->> C)What is the magnitude of the acceleration of the elevator if the scale reads 686.0 N ?
D)In which direction is the elevator moving if the scale reads 686.0 N ?
Explanation / Answer
Given that :
weight of the physics student, W = 516 N
mass of the elevator, m = 868 kg
(a) magnitude of the acceleration of an elevator at the instant of time when the scale reads 449 N which is given as :
using an equation, Ftotal = W + FN { eq.1 }
where, FN = normal force on elevator = m a
Ftotal = net force on scale = 449 N
then, Ftotal = W + m a { eq.2 }
inserting the values in eq.2,
(449 N) = (516 N) + (868 kg) a
- (67 N) = (868 kg) a
a = - 0.077 m/s2
(b) Tension in the elevator cable in Part (A) will be given as ::
T = m a
T = (868 kg) (-0.077 m/s2)
T = - 66.8 N
When the Scale Reads More than the Weight of the Student.
(c) magnitude of the acceleration of the elevator if the scale reads 686 N which is given as :
using eq.2, Ftotal = W + m a
where, Ftotal = 686 N
inserting the values in eq.2,
(686 N) = (516 N) + (868 kg) a
(170 N) = (868 kg) a
a = (170 N) / (868 kg)
a = 0.19 m/s2
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