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Updated on : 2022-09-05

Solution

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In this problem, the work done is $W=Fd$, where F is the tension in the cord and d is the length of the cord pulled as the cart slides from $x_{1}$ to $x_{2}$.

From the figure, we have

$d=x_{1}+h_{2} −x_{2}+h_{2} $

$=(3.00m)_{2}+(1.20m)_{2} −(1.00m)_{2}+(1.20m)_{2} $

$=3.23m−1.56m=1.67m$

which yields $ΔK=Fd=(25.0N)(1.67m)=41.7J.$

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