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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°, the box starts to slip and slides 4.0 m down the plank in 4.0 s. 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)

As we learnt in

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.

 

 M_{s}=tan\: 30^{\circ}=\: \frac{1}{\sqrt{3}}=\: 0.6

S=ut+\frac{1}{2}at^{2},\ \: \: \: u=0,\ \: \: a=g(sin\theta-Mcos\theta)

4=\frac{1}{2}g (sin 30-M_kcos30)4^{2}

\Rightarrow 0.5\times 10\times \frac{1}{2}-M_k\times 10\times \frac{\sqrt{3}}{2}

\Rightarrow 5\sqrt{3}M_k=4.5\ \: \: \Rightarrow M_k=0.51

Correct option is 1.


Option 1)

0.6 and 0.5

Correct

Option 2)

0.5 and 0.6

Incorrect

Option 3)

0.4 and 0.3

Incorrect

Option 4)

0.6 and 0.6

Incorrect

Posted by

Vakul

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