Q.
A block mass 10 kg is kept on a horizontal surface. A force $F$ is acted on the block as shown in figure. For what minimum value of $F$, the block will be lifter up?
A) 98 N
B) 49 N
C) 200 N
D) None of these
Explanation
(c) $F\sin{30{^\circ} + N = Mg}$
The block lifts when $N = 0$
$\therefore\ F = \frac{10 \times 10}{\frac{1}{2}} = 200$N
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.