# Chapter 7 - Kinetic Energy and Work - Problems - Page 171: 15a

Work Done $=1.50J$

#### Work Step by Step

In order to find net Work Done by the three forces, we need to find the resultant force on the block along the horizontal direction. Since there's no vertical displacement, Work done by any force will be Zero in the vertical direction. But we take into account only the forces $F_{1}$ and $F_{2}$, as $F_{3}$ has no horizontal component. $F_{2}$'s component along the horizontal direction is given by: $F_{2}cos\theta= 9\times cos60^{\circ}= 9/2=4.5N$ So the net force in horizontal direction is: $5N +(-4.5N)=0.5N$ $Work$ $Done= F.d=0.5N\times 3m = 1.50J$

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