A particle moves from the origin in three seg | vedclass

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(A) The work done by a constant force acting in the direction of displacement is given by $W = F \cdot d$,where $d$ is the distance covered.Since the

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$(2, 2, 2) \, m$

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(A) The work done by a constant force acting in the direction of displacement is given by $W = F \cdot d$,where $d$ is the distance covered.Since the force $F = 50 \, N$ acts along the direction of motion in each segment,the work done in each segment is $W_i = F \cdot d$.Given that the total work done is $W_{total} = 300 \, J$ and the work done in each of the three segments is equal,the work done in each segment is $W_i = \frac{300}{3} = 100 \, J$.Using $W_i = F \cdot d$,we find the distance $d$ for each segment: $100 = 50 \cdot d$,which gives $d = 2 \, m$.The particle moves $2 \, m$ along the $x$-axis,then $2 \, m$ parallel to the $y$-axis,and finally $2 \, m$ parallel to the $z$-axis.Starting from the origin $(0, 0, 0)$,the final coordinates are $(2, 2, 2) \, m$.

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