A particle moves from point A (5$$, $$6) to point B (3$$, $$4) under the influence of a constant force F→$$=(2i^+bj^)$$ N. If the speed of the particle remains constant, then the value of ‘b’ (in$$ $$N) is

Displacement, d→$$=(3$$−$$5)i^+(4$$−$$6)j^$$ $$m=-2i^-2j^W$$ $$=$$ work $$done=F$$→.d→$$=2($$−$$2)+b($$−$$2)=($$−$$4$$−$$2b)Since$$ kinetic energy remains constant, $$W=0$$⇒−$$4$$−$$2b=0$$⇒$$b=$$−$$2$$

Work Energy Theorem

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Question

A particle moves from point A (5, 6) to point B (3, 4) under the influence of a constant force $\vec{F} = (2 \hat{i} + b \hat{j}) N$ . If the speed of the particle remains constant, then the value of ‘b’ (in N) is

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Solution

Displacement, $\vec{d} = (3 - 5) \hat{i} + (4 - 6) \hat{j} m = - 2 \hat{i} - 2 \hat{j}$
W = work done $= \vec{F} . \vec{d}$
$=2 (- 2) + b (- 2) = (- 4 - 2 b)$
Since kinetic energy remains constant,
$W = 0 \Rightarrow - 4 - 2 b = 0 \Rightarrow b = - 2$

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