A block of mass m=2 kg is placed on a plank of mass M=10 kg, which is placed on a smooth horizontal plane. The coefficient of friction between the block and the plank is μ=13. If a horizontal force F is applied on the plank, then the maximum value of F in newton for which the block and the plank move together is (g=10 m/s2)

a=F−fM=F−f10→(2)F−f10=f2F−f=5fF=6ff=F6Not to slidef=fsmaxFmax6=μs2gFmax=13×20×6RFmax=40NNot to topplef×h2=N×h8f=N4=mg4=204=5Fmax 6=5Fmax=30NFmax=30N to move blocks together.

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

A block of mass m=2 kg is placed on a plank of mass M=10 kg, which is placed on a smooth horizontal plane. The coefficient of friction between the block and the plank is $μ = \frac{1}{3} .$ If a horizontal force F is applied on the plank, then the maximum value of F in newton for which the block and the plank move together is $(g = 10 m / s^{2})$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 30.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} a = \frac{F - f}{M} = \frac{F - f}{10} \to (2) \\ \frac{F - f}{10} = \frac{f}{2} \\ F - f = 5 f \\ F = 6 f \\ f = \frac{F}{6} \end{array}$
Not to slide
$\begin{array}{l} f = f_{s max} \\ \frac{F_{max}}{6} = μ_{s} 2 g \\ F_{max} = \frac{1}{3} \times 20 \times 6 \\ R_{F_{max}} = 40 N \end{array}$
Not to topple
Question Image
$\begin{array}{l} f \times \frac{h}{2} = N \times \frac{h}{8} \\ f = \frac{N}{4} = \frac{mg}{4} = \frac{20}{4} = 5 \\ \frac{F_{max}^{}}{6} = 5 \\ F_{max} = 30 N \\ F_{max} = 30 N to move blocks together. \end{array}$