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A plank with a box on it at one end is gradually raised about the other end. As the angle of inclination with the horizontal reaches $30^{o}$ the box starts to slip and slides 4.0m down the plank in 4.0s. The coefficients of static and kinetic friction between the box and the plank will be, respectively:

Option 1)
0.6 and 0.5

Option 2)
0.5 and 0.6

Option 3)
0.4 and 0.3

Option 4)
0.6 and 0.6

Answers (1)

best_answer

As we discussed in concept

Angle of Friction -

Angle $heta$ is called angle of friction.

$tan heta=frac{F_{l}}{R}$

$tan heta=mu_{s}$

R = Reaction, $frac{F_{l}}{R}=mu_{s}$

$F_{l}=$ Force of limiting friction

$heta=tan^{-1}(mu_{s})$

- wherein

$mu_{s}=$ static friction coefficient

$heta=$ Angle made by resultant of limiting friction and R with the normal reaction.

$\mu _{s}=tan \theta$

$\mu _{s}=tan 30^{\circ}=\frac{1}{\sqrt{3}}=0.577\simeq 0.6$

$S=ut+\frac{1}{2}at^{2}$

$4=\frac{1}{2}a(4)^{2}\:\:\:=a=\frac{1}{2}=0.5\:ms^{-2}$

$a=g sin \theta - \mu_{k}\:g cos \theta$

$\therefore\: \mu_{k}= \frac{0.9}{\sqrt{3}}=0.5$

Option 1)

0.6 and 0.5

This option is correct.

Option 2)

0.5 and 0.6

This option is incorrect.

Option 3)

0.4 and 0.3

This option is incorrect.

Option 4)

0.6 and 0.6

This option is incorrect.

Posted by

prateek

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