An elevator starts from rest with a constant
upward acceleration and moves 1m in the first
1.5 s. A passenger in the elevator is holding a
9.2 kg bundle at the end of a vertical cord.
The acceleration of gravity is 9.8 m/s^2.
What is the tension in the cord as the ele-
vator accelerates? Answer in units of N.
2006-12-20 12:17:20 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
You don't multiply g-force by 1.5 (per previous answer). You add g-force to elevator acceleration force, or just add the accelerations and then compute force. Under constant acceleration,
accel=2*distance/time^2 = 2/2.25 = .8889
accel(total) = .8889 + 9.8
tension = m * accel(total) = 9.2*(.8889 + 9.8) = 98.34 N
2006-12-20 12:25:04 · answer #1 · answered by kirchwey 7 · 0⤊ 0⤋
9.2Kg * 9.8m/s^2 = 90.16N
90.16N * 1.5 = 135.2N
answer = 135.2N
2006-12-20 20:22:46 · answer #2 · answered by sportsfreak 2 · 0⤊ 0⤋