physics power problem explanation, please!?
2020-07-22 1:45 pm
A 2.00 kg object is accelerated uniformly from rest to 3.00 m/s while moving 1.5 m across a level frictionless surface. Calculate the power output.
回答 (3)
2020-07-22 2:14 pm
Initial velocity, u = 0 m/s
Final velocity, v = 3.00 m/s
Displacement, s = 1.5 m
s = [(u + v)/2] t
1.5 = [(0 + 3.00)/2] t
Time taken, t = 1 s
Power output, P
= (Work done) / t
= (Increase in K.E) / t
= (1/2) m v² / t
= (1/2) (2) (3.00)² / (1)
= 9 W
Final velocity, v = 3.00 m/s
Displacement, s = 1.5 m
s = [(u + v)/2] t
1.5 = [(0 + 3.00)/2] t
Time taken, t = 1 s
Power output, P
= (Work done) / t
= (Increase in K.E) / t
= (1/2) m v² / t
= (1/2) (2) (3.00)² / (1)
= 9 W
2020-07-22 4:16 pm
You can calculate the AVERAGE power output but not THE power output because the output is continually increasing. At each instant the power is F v but we know F = ma and a = vf^2 /(2s) ->P = m v vf^2/(2s)
which is zero initially and rises to a maximum of 2*3 * 3^2 / ( 2 * 1.5) = 18W
The AVERAGE power can be found from total energy / total time = 1/2 m v^2 /( s/(v/2) ) = 1/2 mv^3/(2s) = 1* 3^3 /(2*1.5) = 9 W Which is half of the maximum power used at the end.
which is zero initially and rises to a maximum of 2*3 * 3^2 / ( 2 * 1.5) = 18W
The AVERAGE power can be found from total energy / total time = 1/2 m v^2 /( s/(v/2) ) = 1/2 mv^3/(2s) = 1* 3^3 /(2*1.5) = 9 W Which is half of the maximum power used at the end.
2020-07-22 3:29 pm
energy KE = m/2*Vf^2 = 1*3^2 = 9.0 joule
acceleration time = 2d/V = 2*1.5/3 = 1 sec
aver. power P = KE/t = 9/1 = 9 watt
acceleration time = 2d/V = 2*1.5/3 = 1 sec
aver. power P = KE/t = 9/1 = 9 watt
收錄日期: 2021-04-18 18:34:44
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