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4-2.Friction
medium
A block of mass $5\, kg$ is kept on a rough horizontal floor. It is given a velocity $33\, m/s$ towards right. A force of $20\sqrt {2\,} \,N$ continuously acts on the block as shown in the figure. If the coefficient of friction between block and floor is $0.5$ the velocity of block after $3\, seconds$ is ........ $m/s$ ($g = 10\, m/s^2$)
A$2$
B$0$
C$33/12$
DNone of above
Solution
Net retarding force $=\mathrm{F} \cos 45^{\circ}+\mu \mathrm{N}$
$=\mathrm{F} \cos 45^{\circ}+\mu\left(\mathrm{F} \sin 45^{\circ}+\mathrm{mg}\right)$
$=20 \sqrt{2} \times \frac{1}{\sqrt{2}}+0.5\left(20 \sqrt{2} \times \frac{1}{\sqrt{2}}+5 \times 10\right)$
$=55 \mathrm{N}$
retardation $=\frac{\mathrm{F}_{\mathrm{Net}}}{\mathrm{m}}=\frac{55}{5}=1 \mathrm{m} / \mathrm{s}^{2}$
$\mathrm{V}=\mathrm{u}-\mathrm{at}=33-11 \times 3=0$
$=\mathrm{F} \cos 45^{\circ}+\mu\left(\mathrm{F} \sin 45^{\circ}+\mathrm{mg}\right)$
$=20 \sqrt{2} \times \frac{1}{\sqrt{2}}+0.5\left(20 \sqrt{2} \times \frac{1}{\sqrt{2}}+5 \times 10\right)$
$=55 \mathrm{N}$
retardation $=\frac{\mathrm{F}_{\mathrm{Net}}}{\mathrm{m}}=\frac{55}{5}=1 \mathrm{m} / \mathrm{s}^{2}$
$\mathrm{V}=\mathrm{u}-\mathrm{at}=33-11 \times 3=0$
Standard 11
Physics
