power problem physics?
2016-03-30 1:34 am
A 2300 kg elevator is attached to a 1800kg counterweight. What power must the motor supply to raise the elevator at 0.4m/s?
回答 (3)
2016-03-30 1:41 am
✔ 最佳答案
P = W/t = (F*d)/t = F*vP = F*v
F = m*g
P = m*g*v
P = (500*9.8*.4) = 1960 W
2016-03-30 1:41 am
Power, P
= F v
= m g v
= (2300 - 1800) × 9.81 × 0.4
= 1962 W
= F v
= m g v
= (2300 - 1800) × 9.81 × 0.4
= 1962 W
2016-03-30 1:57 am
Effective mass of Elevator = 2300 - 1800 = 500 kg
initial velocity, u = 0
final velocity, v = 0.4 m/s
Power = change in Kinetic Energy
P = ΔKE
P = 1/2 mv² - 1/2 mu²
P = 1/2 (500)(0.4)² - 0
P = 40 W
initial velocity, u = 0
final velocity, v = 0.4 m/s
Power = change in Kinetic Energy
P = ΔKE
P = 1/2 mv² - 1/2 mu²
P = 1/2 (500)(0.4)² - 0
P = 40 W
收錄日期: 2021-04-18 14:41:05
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160329173433AAqtE8d