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A $1.8 kg$ block is moved at constant speed over a surface for which coefficient of friction $μ = \frac{1}{4}$ . It is pulled by a force F acting at $45 °$ with horizontal as shown in Fig. The block is displaced by $2 m$ . Find the work done on the block by friction.

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-6.2 J

-7.2 J

-3.2 J

-5.2 J

answer is C.

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Detailed Solution

$W_{f} = (μ N) (S) \cos 180 ° = (\frac{1}{4}) (18 - \frac{F}{\sqrt{2}}) (2) (- 1) = - 7.2 J$

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