JEE Main 2018 · Physics · Kinematics and Laws of Motion · medium

A smooth wedge of mass 6 kg with an inclined face at 37^° to the horizontal rests on a smooth horizontal floor. A block of mass 2 kg lies on the…

Single correct option

A smooth wedge of mass

6\ \mathrm{kg}

with an inclined face at

37^\circ

to the horizontal rests on a smooth horizontal floor. A block of mass

2\ \mathrm{kg}

lies on the inclined face. A horizontal force

F

pushes the wedge so that the block stays at rest relative to the wedge. Taking

g=10\ \mathrm{m\,s^{-2}}

and

\tan37^\circ=\tfrac34

, the value of

F

Show answer and solution

Answer: (A) $60\ \mathrm{N}$

The block moves with the wedge, with common acceleration

a

horizontally. Its only forces are gravity and the normal reaction

N

perpendicular to the smooth face: vertical balance gives

N\cos37^\circ=mg

, and horizontal acceleration gives

N\sin37^\circ=ma

. Dividing,

a=g\tan37^\circ=7.5\ \mathrm{m\,s^{-2}}

. Treating wedge and block as one body on the smooth floor,

F=(M+m)a=8\times7.5=60\ \mathrm{N}

. Using

a=g\tan53^\circ

by mistake gives

107\ \mathrm{N}

, and

Ma

alone gives

45\ \mathrm{N}

Original question written in the JEE Main 2018 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake