A box weighing 2000N is to be slowly slid through 20m 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 'theta' with the horizontal. (b) Find the work one when the person has chosen a value of 'theta' which ensure him the minimum magnitude of the force.
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(a) Suppose F is the force which is applied by the person at an angle θ with the horizontal. Let N be the normal force of the track on the box.Then,N=mg−FsinθFcosθ=f=μ(mg−Fsinθ)So,F=μmgcosθ+μsinθNow, Work done, W=Fdcosθ=μmgcosθ+μsinθ×dcosθ=μmgd1+μtanθ(Dividing numerator and denominator by cosθ)Putting the values,W=0.2×2000×201+0.2tanθ=400005+tanθJ(b) From the above, we haveF=μmgcosθ+μsinθThe force will be minimum if cosθ+μsinθ is maximum.And cosθ+μsinθ is maximum if it's first derivative is zero.So, −sinθ+μcosθ=0or tanθ=μ=0.2Thus, W=400005+tanθ=400005+0.2=7690 J
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