Answer to Question #6294 in Mechanics | Relativity for ian luna
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 .
Using the Newton'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'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.