Work and Energy

as in all cases the block is pulled solwly
it always remains in equillibrium

in the first case let the man apply a force F at an angle θ to the horizontal

then as it always remains in equillibrium

Fcosθ = μN

but N=mg-Fsinθ
so

F= μmg/(cosθ+μsinθ)

but wrok done by the man =
F.ds = FLcosθ = L(μmgcosθ)/(cosθ+μsinθ)

replacing the values we get answer as

40000/(5+tanθ)

for the second case

we had already got

F=μmg/(cosθ+μsinθ)

now, F will be minimum when the denominator will be maximum

but max. value of (cosθ+μsinθ) = √(μ²+1)

this will be when cosθ = 1/√(μ²+1)

hence work done =

μmgLcosθ/√(μ²+1)

putting the values given in ques we get ans as 7690 J