Asked by Margaret
Hi! I'm supposed to derive an algebraic formula to solve for the acceleration using one of the linear kinematics formulas of an elevator with a person inside that begins to accelerate from rest moving downward. If it travels a know distance, d, in reaching its max downward speed( unknown values). Assume you have access to a timekeeping device.
So I derived acceleration to be a= (v^2)/2d
Then it asks me if I know the total mass of the elevator and the person inside, how would you determine the net force acting on the elevator?
someone please please let me know if I'm doing the first part correct and please help me with the second part! Thank you!!
So I derived acceleration to be a= (v^2)/2d
Then it asks me if I know the total mass of the elevator and the person inside, how would you determine the net force acting on the elevator?
someone please please let me know if I'm doing the first part correct and please help me with the second part! Thank you!!
Answers
Answered by
Damon
mass = m
F is our unknown force up (maybe from cable)
a is constant and negative (down)
there is a constant downward force, the weight, mg
it starts at x = 0, and Vi = 0 (initial speed)
F - mg = - m d^2x/dt^2 note F is up, x is down
-m dx/dt = - m v = - m a t
so
v = a t + a constant which is zero because v = 0 at t = 0
so t = v/a
then
x = integral v dt = a t^2/2 + constant which again is zero
so
x = a (v^2/a^2)/2 = (1/2)v^2/a
or as you said
a = v^2/(2d)
F- m g = -m a
F = m g - m a = m (g - v^2/2d)
F is our unknown force up (maybe from cable)
a is constant and negative (down)
there is a constant downward force, the weight, mg
it starts at x = 0, and Vi = 0 (initial speed)
F - mg = - m d^2x/dt^2 note F is up, x is down
-m dx/dt = - m v = - m a t
so
v = a t + a constant which is zero because v = 0 at t = 0
so t = v/a
then
x = integral v dt = a t^2/2 + constant which again is zero
so
x = a (v^2/a^2)/2 = (1/2)v^2/a
or as you said
a = v^2/(2d)
F- m g = -m a
F = m g - m a = m (g - v^2/2d)
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