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Camping equipment weighing 6000n is pulled across a frozen lake by means of a horizontal rope. the coefficient 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/s2

Sagot :

Answer:

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