Q. A box weighing 2000 N is to be slowly slid through 20 m on a straight track having friction coefficient 0.2 with the box. (a) Find the work done by the person pulling the box with a chain at an angle θ with the horizontal. (b) Find the work when the person has chosen a value of θ which ensures him the minimum magnitude of the force.
[Ans. (a) 40000 J/ 5+tanθ (b) 7690 J ]
Plz explain the answer.
1as 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)
1for 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
Anonymous how is costheta +usintheta max value root of u^2+1