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 i
s holding a 5.1 kg bundle at the end of a vertical cord. What is the tension in the cord as the elevator accelerates
1 answer:
Answer: Tension = 53.6N
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
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