- #1
get_rekd
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A 700 kg elevator suspended by a cable accelerates upwards at 3 m/s^2. The force exerted by the cable on the elevator is?
(use g=10 m/s^2)
m=700 kg
a= 3.0 m/s^2
Fnet= 700 kg * 3.0 m/s^2
= 2,100.0 N Upward force
Fnet= 700 kg * 10 m/s^2
= 7,000.0 N Downward force
Fnet= 2,100.0 N + 7,000.0 N = 9,100.0 N
So is 9,100.0 N the total force exerted on the cable?
(use g=10 m/s^2)
m=700 kg
a= 3.0 m/s^2
Fnet= 700 kg * 3.0 m/s^2
= 2,100.0 N Upward force
Fnet= 700 kg * 10 m/s^2
= 7,000.0 N Downward force
Fnet= 2,100.0 N + 7,000.0 N = 9,100.0 N
So is 9,100.0 N the total force exerted on the cable?