Question

Camping equipment weighing 6000 n 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 1000 m if its speed is increasing at the constant rate of 0.20 m/s2?

Answer 1

Hello Dear
Your Answer is :

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 
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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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Answer 2

Answer:

4.2×10^5 is the total work done calculated by adding work against friction and work by force