Circular track

At the bottom of the track, the normal force on the block is upward, gravity acts downward, and the net centripetal force is upward, so $n - mg = \frac{mv^2}{R}$ . Plugging in the given values, we find that $v = \sqrt{\frac{R(n-mg)}{m}} = \sqrt{\frac{(0.8)(3.40-(0.05)(9.8))}{0.05}}\approx 6.8235$ m/s. Using conservation of energy between the bottom and top parts of the loop, we know that $\frac{1}{2}mv^2 = mg\Delta y + \frac{1}{2}mv_{top}^2$ where $\Delta y = 2R$ : $\frac{1}{2}v^2 = 2gR + \frac{1}{2}v_{top}^2$ $v_{top} = \sqrt{2(\frac{1}{2}v^2 - 2gR)} = \sqrt{(6.8235)^2 - 4(9.8)(0.8)} \approx 3.899 m/s$ At the top of the loop, the normal force on the block acts downward, so $n + mg = \frac{mv_{top}^2}{R}$ . Now we find that $n = \frac{mv_{top}^2}{R} - mg = \frac{(0.05)(3.899)^2}{0.8} - (0.05)(9.8) \approx \boxed{0.46 N}$ .

$$ Let $\,N_T$ and $\,N_B$ be the normal force at the top and bottom of circular track, respectively. Let $\,v_T$ and $\,v_B$ be the normal force at the top and bottom of circular track, respectively. At the bottom of circular track, applying the Newton's $\,2^{nd}$ law for circular motion. $$ $\begin{aligned} N_B - mg &= m\frac{v_B^2}{R}\\ mv_B^2&= (N_B - mg)R, &&&&& (1) \end{aligned}$ $$ then use the law of conservation energy for the movement of the block from the bottom to the top of circular track and plug in equation (1). We obtain $$ $\begin{aligned} \frac{1}{2}mv_B^2&=mg(2R)+\frac{1}{2}mv_T^2\\ \frac{1}{2}(N_B - mg)R&=2mgR+\frac{1}{2}mv_T^2\\ mv_T^2&=(N_B - 5mg)R.&& (2) \end{aligned}$ $$ Finally, use the Newton's $\,2^{nd}$ law for circular motion at the top of circular track and plug in equation (2). We obtain $$ $\begin{aligned} N_T + mg &= m\frac{v_T^2}{R}\\ N_T + mg &=N_B - 5mg\\ N_T &=N_B - 6mg\\ N_T &=3.4-6\cdot 0.05\cdot 9.8\\ N_T &=\boxed{0.46\;\text{N}} \end{aligned}$ $$ $\text{\# }\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}$

Let linear velocity of the block at the bottom and at the top be u and v respectively.
Therefore, total energy of the block at the bottom = $\frac{1}{2}$ m $u^{2}$ + $V_{1}$
and that of the block at the top = $\frac{1}{2}$ m $v^{2}$ + $V_{2}$ , where m =mass of the block and $V_{1}$ and $V_{2}$ are potential energies at the bottom and top respectively.
Applying law of conservation of energy,
$\frac{1}{2}$ m $u^{2}$ + $V_{1}$ = $\frac{1}{2}$ m $v^{2}$ + $V_{2}$
or, m ( $u^{2}$ - $v^{2}$ ) = 2( $V_{2}$ - $V_{1}$ ) or, m ( $u^{2}$ - $v^{2}$ )=2 mg .(2R) or, m ( $u^{2}$ - $v^{2}$ )= 4 mg R
or, $v^{2}$ = $u^{2}$ - 4 g R .....................(i)
Now, considering forces acting at the bottom of the path, we get,
$\frac{mu^{2}}{R}$ + m g = 3.40 or, $\frac{mu^{2}}{R}$ = 3.4 - mg ........................(ii)
and considering forces acting on the top of the path, we get, P = $\frac{mv^{2}}{R}$ - m g , where P = required normal force.
Using (i) and (ii), we get, P = 3.4 - 6 mg
Putting m = 0.0500 kg and g = 9.8 m/ $s^{2}$ we get,
P = $\boxed{0.46 N}$

At the top of the circle the normal force exerted by the track will be opposite to that of the sliding block; that is subject to two forces: gravity and the centrifugal force.

Hence we have

$F=m\times a_s= m(\frac{1}{R}{V_h }^2-g)$

where ${V_h }$ represents the speed in the block highest position while at the bottom we know

$F_l=3.4N=m(\frac{1}{R}{V_l }^2+g)$

Subtracting the first equation from the last one we get

1) $F_l-F=\frac{m}{R}({V_l }^2-{V_h}^2) +2mg$

On the other hand, the difference in kinetic energy equals the potential one, between top and bottom, so

$\frac{1}{2}m({V_l }^2-{V_h}^2)=mg\times2R$

that is

${V_l }^2-{V_h}^2=4gR$

Then we can substitute it in the 1), obtaining at last

$F=F_l-4mg-2mg=F_l - 6g$