A block of mass $5 k g$ is initially at rest on a rough horizontal surface. A force of $45 N$ acts on it in a horizontal direction and pushes it over a distance of $2 m$ . The force of friction acting on the block is $25 N$ . The final kinetic energy of the block is

40 J

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90 J

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50 J

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140 J

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Solution

The correct option is A $40 J$
Using work-energy theorem:
$W . D_{f o r c e} + W . D_{f r i c t i o n} = Δ K . E$
$45 \times 2 - 25 \times 2 = K E_{f i n a l}$
$K E_{f i n a l} = 40 J$

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