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When a force of 10 N acts on a particle at 37° east of north then produces a displacement of 5 m South of West. Find work done.

Eshan
3 years ago
Dear student,

Angle between displacement and force vectors=$\dpi{80} 90^{\circ}+45^{\circ}+37^{\circ}=172^{\circ}$

Hence work done=$\dpi{80} Fscos\theta=10\times 5\times cos172^{\circ}$

$\dpi{80} cos172^{\circ}=cos(180^{\circ}-8^{\circ})=-cos8^{\circ}=-cos(45^{\circ}-37^{\circ})$

$\dpi{80} =-(cos45^{\circ}cos37^{\circ}+sin45^{\circ}sin37^{\circ})=-\dfrac{1}{\sqrt{2}}$

Hence work done=$\dpi{80} -25\sqrt{2}J$