Help with the solution please
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Help with the solution please

[From: ] [author: ] [Date: 12-02-18] [Hit: ]
pulling the equipment 1000m if its speed is increasing at the constant rate of 0 .20m /s^2?-Acceleration a = 0.Actual acceleration a=0. 2 + μg =0. 2 0+ 0.......
Camping equipment weighing 6000N is pulled across a frozen lake by means of a horizontal
rope. The coe fficient of kinetic friction is 0.05. How much work is done by the campers in
pulling the equipment 1000m if its speed is increasing at the constant rate of 0 .20m /s^2?

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Acceleration a = 0. 20 m/s^2
Actual acceleration a'= 0. 2 + μg = 0. 2 0+ 0.05*9.8 = 0.69 m/s^2

work done = force * distance = m a' s = ( 6000/9.8) *0.69 *1000 =422449 J
=======================================…

Or


Wokdone m a 1000 = F*a/g = 6000 *(0.2/9.8)*1000 = 122.449
Work aginst friction = 6000*0.05*1000 =300000

Total work done = 422.5*1000 = 422.449
=======================================…

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work done is W = uk m g S = 0.05 x 6000 x 1000 = 300 KJ.
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