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# Answer to Question #6294 in Mechanics | Relativity for ian luna

Question #6294
An elevator starts from rest with a constant
upward acceleration andmoves 1m in the first
1.9 s. A passenger in the elevator is holding a
7.4 kg bundle at the end of a vertical cord.
What is the tension in the cord as the ele-
vator accelerates? The acceleration of gravity
is 9.8 m/s2 .
1
2012-02-09T08:25:07-0500
Using the Newton&#039;s Second Law the tension can be found from the
equation:
Tension = F = ma, where a is the sum of the acceleration of gravity
and the acceleration of elevator:
a = g + aElev.
As the elevator moves
with a constant acceleration, let&#039;s write an equation for displacement:
S =
V0*t + (aElev*t^2)/2. As the initial speed V0 = 0, then:
S = (aElev*t^2)/2.
So, aElev = 2S/t^2. Consecuently, We get the formula for tension:
T = m(g +
2S/t^2) = 7.4*(9.8+2*1/1.9^2) = 76.6 N.

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