# A block of weight $W$ is pulled along a distance $d$ on a table. The work done by the weight is

$(a)\;Wd\quad (b)\;0 \quad (c)\;Wgd \quad (d)\;Wd/g$

Since the work done on the weight is perpendicular to the direction of displacement $\theta = 90^\circ$
Work done = $F.d \cos \theta$. Since $\cos \theta = 0$, work done $=0$
edited Aug 25, 2014