Answer:
Work done = 422.45 kJ
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'
[tex]F = \dfrac{6000}{9.8} \times 0.69[/tex]
F = 422.44897 N
Work done = F x d
Work done = 422.44897 x 1000
Work done = 422449 J
Work done = 422.45 kJ