Question

An elevator starts from rest with a constant upward acceleration and moves 1 m in the first 1.7 s. A passenger in the elevator is holding a 5.1 kg bundle at the end of a vertical cord. What is the tension in the cord as the elevator accelerates

Answer 1

Answer: Tension = 53.6N

Step-by-step explanation:

Given that

Height h = 1 m

Time t = 1.7 s.

Mass m = 5.1 kg

From the equation of the motion we can get the acceleration of the elevator:

h = X0+ V0t + at2/2;

Th elevator starts from rest with a constant upward acceleration. Initial velocity Vo = 0, also Xo = 0; thus

a = 2h/t2 = 2 × 1/1.7^2

a = 0.69 m/s2.

Then we can find the tension in the cord by using the formula

T = mg + ma

= 5.1 (9.8 + 0.69)

= 5.1 × 10.5

= 53.6N