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A student stands on a bathroom scale in an elevator at rest on the 53th floor of a building. ?
Question : A student stands on a bathroom scale in an elevator at rest on the 53th floor of a building. ?
The scale reads 867 N.
As the elevator moves up, the scale reading increases to 991 N, then decreases back to 867 N. Find the acceleration of the elevator.
As the elevator approaches the 63th floor, the scale reading drops to 683 N. What is the acceleration of the elevator?
floor scale
Best answer:
Answer by Blair A
using Newtons 2nd law of motion we find that F(unbalanced) = ma
Weight = mg
mass = weight/gravitational field strength
mass = 867/9.81(assuming hes on earth)
mass = 88.4kg.
F(unbalanced) = 991 – 867 = 124 N
124 = 88.4 x a
a = 1.4 m/s/s
same again for part b)
mass = 88.4kg
Fub = 683 – 867 = -184 (- implies direction)
a = -2m/s/s or deceleration by 2 m/s/s whatever u want
Cheers
Blair
This question is quite simple:
First find the mass of the student:
F = m * g where F is Force, M is mass, and g is gravity (assume 9.81m/s^2)
m = F/g
m = 867N / 9.81 m/s^2 = 88.4 kg
For a)
We need to find the acceleration of the elevator. The upward force is 991 (to lift the elevator up) and the downward force is 867N (from the student in the elevator). So the difference is UPWARD – DOWNWARD; 991N – 867N = 124N.
We know that F = ma (Newtons law). F being the difference of the upward and downward forces (completed above).
a = F/m = 124N / 88.4 kg = 1.40 m / s^2
That means the elevator is accelerating as its moving up.
b) Again the difference of the UPWARD and DOWNWARD forces is 683N – 867 N = -184N.
Using Newton’s Law again:
F = ma
a = F / m
a = -184N / 88.4 kg = -3.5 m / s^2
That means that as it is approaching the 63th floor, it starts to decelerate at a rate os 3.5 m / s^2