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A::CSolution :

In this case, three forces are acting on the object: <br> 1. tension (T) <br> 2. weight (mg) and <br> 3. applied force (F) <br> Using work-enegy theorem <br> (##DCP_V01_C09_S01_010_S01##). <br> `W _ (n et)=DeltaKE` <br> or `W_(T) + W_(mg) + W_(F)=0` ...(i) <br> as `DeltaKE=0` <br> because `K_(i) = K_(f) = 0` <br> Further, `W_T=0,` . as tension is always perpendicular to displacement. <br> `W_(mg) =-mgh` <br> or `w_(mg)=-mgl(1-costheta)` <br> Substituting these values in Eq. (i), we get <br> `W_(F) = mgl(1-costheta)`.