Let's analyze the forces acting on the box:
The component of weight parallel to the incline is mg sin30° = 50 × $\frac{1}{2}$ = 25 N (down the incline).
Applying Newton's second law along the incline (upward direction as positive):
T − mg sin 30° = ma
30 N − 25 N = 5 kg × a
a = $\frac{5}{5}$ = 1 m s-2
A body of mass 1kg is suspended with the help of two strings making angles as shown in the figure. Magnitude of tensions $ T_1 $ and $ T_2 $, respectively, are (in N):
A bob of heavy mass \(m\) is suspended by a light string of length \(l\). The bob is given a horizontal velocity \(v_0\) as shown in figure. If the string gets slack at some point P making an angle \( \theta \) from the horizontal, the ratio of the speed \(v\) of the bob at point P to its initial speed \(v_0\) is :
A full wave rectifier circuit with diodes (\(D_1\)) and (\(D_2\)) is shown in the figure. If input supply voltage \(V_{in} = 220 \sin(100 \pi t)\) volt, then at \(t = 15\) msec: