Answer:
Work done = 422.45 kJ
Step-by-step explanation:
given, 
weight of equipment = 6 kN 
coefficient of kinetic friction = 0.05 
distance up to which it is pulled = 1000 m
constant acceleration = 0.2 m/s² 
Work done by the camper = ? 
actual acceleration acting a' 
 m a = m a' - μ mg 
a' = a + μ g 
a' = 0.2 + 0.05 x 9.8 
a' = 0.69 m/s² 
Work done = Force x distance 
F = m a' 

F = 422.44897 N 
Work done = F x d 
Work done = 422.44897 x 1000 
Work done = 422449 J 
Work done = 422.45 kJ