a block of mass m is released from rest at a height r above a horizontal surface. the acceleration due to…

Explanation:

Step1: Apply conservation of mechanical energy

Initial mechanical energy $E_1 = mgh$, with $h = R$ (initial height), so $E_1=mgR$. At the bottom of the hoop, potential energy is zero and kinetic energy is $K=\frac{1}{2}mv^{2}$, and total mechanical energy $E_2=\frac{1}{2}mv^{2}$. By conservation of energy $E_1 = E_2$, so $mgR=\frac{1}{2}mv^{2}$.

Step2: Solve for velocity

From $mgR=\frac{1}{2}mv^{2}$, we can solve for $v$. Multiply both sides by 2 and divide by $m$ to get $v^{2} = 2gR$.

Step3: Apply Newton's second - law at the bottom of the hoop

At the bottom of the hoop, the net - force towards the center is $F_{net}=N - mg$, where $N$ is the normal force and $mg$ is the weight. According to Newton's second - law $F_{net}=ma_c$, and the centripetal acceleration $a_c=\frac{v^{2}}{R}$. So $N - mg=\frac{mv^{2}}{R}$.

Step4: Substitute $v^{2}$ into the Newton's second - law equation

Substitute $v^{2}=2gR$ into $N - mg=\frac{mv^{2}}{R}$. We have $N - mg=\frac{m\times2gR}{R}=2mg$. Then add $mg$ to both sides to solve for $N$, $N=2mg + mg=3mg$.

Answer:

d. 3mg