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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.

Draco , 6 Years ago
Grade 12th pass
anser 1 Answers
Eshan

Last Activity: 6 Years ago

Dear student,

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

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

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

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

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

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