A block of mass 25 kg is pulled along a horizontal surface by a force at an angle $45^{\circ}$ with

Question

A block of mass 25 kg is pulled along a horizontal surface by a force at an angle $45^{\circ}$ with the horizontal. The friction coefficient between the block and the surface is 0.25 . The block travels at a uniform velocity. The workdone by the applied force during a displacement of 5 m of the block is\

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(c) : $N+F \sin 45^{\circ}=m g \Rightarrow N=m g-\frac{F}{\sqrt{2}}$\ As the block is travelling with uniform velocity, so acceleration is zero.\ i.e., $a=0 \Rightarrow F_{	ext {net }}=0$\ $F \cos 45^{\circ}=\mu \mathrm{N}$\ $\frac{F}{\sqrt{2}}=0.25\left[m g-\frac{F}{\sqrt{2}}ight]$\ $\frac{F}{\sqrt{2}}=0.25\left[25 	imes 9.8-\frac{F}{\sqrt{2}}ight]$\ $F=49 \sqrt{2} \mathrm{~N}$\ $W_{	ext {ext }}=F \cos 45^{\circ} 	imes d$\ $W_{	ext {ext }}=F 	imes 5 \cos 45^{\circ}=49 \sqrt{2} 	imes 5 	imes \frac{1}{\sqrt{2}}=245 \mathrm{~J}$\