A block of mass 25 kg is pulled along a hori | vedclass

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(C) Since the block travels with uniform velocity,the net force on it is zero $(a = 0)$.Resolving forces horizontally: $F \cos 45^{\circ} = f_k$,where

Vedclass · Accepted Answer

$245$

Vedclass · Answer

(C) Since the block travels with uniform velocity,the net force on it is zero $(a = 0)$.Resolving forces horizontally: $F \cos 45^{\circ} = f_k$,where $f_k = \mu N$.Resolving forces vertically: $N + F \sin 45^{\circ} = mg \Rightarrow N = mg - \frac{F}{\sqrt{2}}$.Substituting $N$ into the friction equation: $\frac{F}{\sqrt{2}} = 0.25 \left( 25 	imes 9.8 - \frac{F}{\sqrt{2}} ight)$.$\frac{F}{\sqrt{2}} = 0.25 	imes 245 - 0.25 \frac{F}{\sqrt{2}}$.$1.25 \frac{F}{\sqrt{2}} = 61.25$.$F = \frac{61.25 	imes \sqrt{2}}{1.25} = 49 \sqrt{2} \ N$.The work done by the applied force is $W = F S \cos 45^{\circ}$.$W = (49 \sqrt{2}) 	imes 5 	imes \frac{1}{\sqrt{2}} = 49 	imes 5 = 245 \ J$.

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