An elevator can carry a maximum load of 1800kg, is moving up with a constant speed of 2m/s. The frictional force opposing the motion is 4000N. Determine the minimum power delivered by the motor to the elevator in watts. (take g=9.8m/s2
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power=p
force=f
t=time
s=displacement
v= velocity
w=work
p=w/t
p=f*s/t
p=f*v
f=1800*9.8 +4000. (m*g+4000 because mg is the minimum force to overcome come earth's gravity and 4000 to overcome force of friction)
f=21640N
p=21640*2
p=43280 watts
force=f
t=time
s=displacement
v= velocity
w=work
p=w/t
p=f*s/t
p=f*v
f=1800*9.8 +4000. (m*g+4000 because mg is the minimum force to overcome come earth's gravity and 4000 to overcome force of friction)
f=21640N
p=21640*2
p=43280 watts
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