Q.
A body of mass 5 kg is suspended by a spring balance on an inclined plane as shown in figure. The spring balance, measure
A) 50 N
B) 25 N
C) 500 N
D) 10 N
Explanation
(b) Acceleration of the body down the rough inclined plane $= g\sin\theta$
$\therefore$ Force applied on spring balance
$= mg\sin{\theta = 5 \times 10 \times \sin{30{^\circ}}}$
$= 5 \times 10 \times \frac{1}{2} = 25\ N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.