An elevator starts from rest with a constant upward acceleration and moves 1 m in the first 1.9 s. A passenger in the elevator is holding a 5 kg bundle at the end of a vertical cord.
What is the tension in the cord as the elevator accelerates? The acceleration of gravity is 9.8 m/s².
Answer in units of N
Solution:
Let:
h=1mt=1.9sm=5kgg=9.8m/s2
F-?
F=mg+ma, were a-acceleration of elevator
h=21at2=2a=2h/t2F=m(g+2h/t2)F=5(9.8+2⋅1/1.92)=51.77N
Answer: 51.77N