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(B) Since the pulley is smooth and light,and the string is light,the tension $T$ throughout the string is equal to the applied force $F$. Thus,$T = F$

Vedclass · Accepted Answer

The tension in the string is $F$.

Vedclass · Answer

(B) Since the pulley is smooth and light,and the string is light,the tension $T$ throughout the string is equal to the applied force $F$. Thus,$T = F$.According to the work-energy theorem,the net work done on the block is equal to the change in its kinetic energy.Net work done $W_{	ext{net}} = \Delta K = 20\,J$.The forces acting on the block are the tension $T$ (upwards) and gravity $Mg$ (downwards).The work done by tension is $W_T = T \cdot d$ and the work done by gravity is $W_g = -Mg \cdot d$,where $d$ is the displacement.$W_{	ext{net}} = W_T + W_g = (T - Mg)d = 20\,J$.Since $T = F$,the work done by the tension is $W_T = F \cdot d$. Since $F > Mg$ (as the kinetic energy increases),$W_T$ must be greater than $20\,J$ because $W_T = 20\,J + Mg \cdot d$.Therefore,the work done by the tension is not $20\,J$,but $20\,J$ plus the work done against gravity. Thus,option $B$ is the correct statement regarding the tension.

$A$ block of mass $M$ is hanging over a smooth and light pulley through a light string. The other end of the string is pulled by a constant force $F$. The kinetic energy of the block increases by $20\,J$ in $1\,s$.

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