physic 難題 4

(a) Total distance travelled by lift = area under the trapezium
= (3+6.5) x 4/2 m = 19 m

(b) Accelration of the lift = 4/(5-3) m/s2 = 2 m/s2
Decceleration of the lift = 4/(9.5-8) m/s2 = 2.67 m/s2

The tension is 15000 N from t = 0 to 3 s
During acceleration, the tension = (1500 x 2 + 1500g) N = 18000 N
At constant speed from t = 5 to 9.5 s, tension = 15000 N
During decceleration, tension = (1500g - 1500 x 2.67) N = 10995 N
From t = 9.5 to 11 s, thension = 15000 N

(c) Use work done = change of kinetic energy
F x 7.5 = (1/2) x 1500 x 12^2
where F is the net force
hence , F = 14400 N