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 .

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 .

Expert's answer

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.

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.

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