JEE Advanced · 2006 · 09 Apr 2006 · Laws of Motion

The block experiences forces from the disc's acceleration and friction from the groove surfaces.

Normal force from bottom surface: $N_1=mg=1\times 10=10$ N

Normal force from side surface: $N_2=ma\sin 37°=1\times 25\times 0.6=15$ N

Friction forces: $f_1=\mu N_1=0.4\times 10=4$ N and $f_2=\mu N_2=0.4\times 15=6$ N

Net force on block in the direction along the groove (in disc frame):

$F_{net}=ma\cos 37°-f_1-f_2=1\times 25\times 0.8-4-6=20-10=10$ N

Acceleration of block with respect to disc:

$a_{bd}=\frac{F_{net}}{m}=\frac{10}{1}=10$ m/s${}^2$

Answer: 10 m/s²