Answer
2.5 N
Work Step by Step
At point Q, the block is kept in the loop by a centripetal force of magnitude $F = \frac{mv^{2}}{R}$ and directed horizontally to the left towards the center of the loop. This will be the horizontal component of the net force on the object.
The problem gives us the mass $m$ = 0.032 kg and the radius $R$ = 12 cm = 0.12 m, but to find the velocity $v$ we use the equation for the conservation of mechanical energy and solve for $v$:
$\Delta E_{mec} = \Delta K + \Delta U = \frac {1}{2} m(v^{2} – v_{0}^{2}) + mg\Delta h = 0$
$\frac {1}{2} m(v^{2} – v_{0}^{2}) = – mg\Delta h$
$v^{2} – v_{0}^{2} = – 2g\Delta h$
$v^{2} = v_{0}^{2} – 2g\Delta h$
$v = \sqrt {v_{0}^{2} – 2g\Delta h}$
And plugging it in we get:
$F = \frac{mv^{2}}{R} $
$= \frac{m(\sqrt {v_{0}^{2} – 2g\Delta h})^{2}}{R}$
$= \frac{m(v_{0}^{2} – 2g\Delta h)}{R}$
The original height of the block was h = 5.0R, and at point Q it is exactly at a height R, thus the change in height $\Delta h$ = 5.0R – R = 4.0R. Since the block is released from rest, $v_{0}$ = 0. Plugging in our values we get:
$F = \frac{m(v_{0}^{2} – 2g\Delta h)}{R}$
$= \frac{m(v_{0}^{2} – 2g(4.0R))}{R}$
$= \frac{(0.032 kg)(0^{2} – 2(9.8 m/s²)(4.0(0.12 m))}{0.12 m}$
= 2.5088 N $\approx$ 2.5 N
