Question

a block of mass m is released from rest at a height r above a horizontal surface. the acceleration due to gravity is g. the block slides along the inside of a frictionless circular hoop of radius r. what is the magnitude of the normal force exerted on the block by the hoop when the block reaches the bottom of the hoop? a. mg²/r b. mg c. 2mg d. 3mg e. zero newtons

Explanation:

Step1: Apply conservation of mechanical energy

The initial mechanical energy of the block is only potential energy given by $E_{i}=mgR$ (since it starts from rest, $v_{i} = 0$). At the bottom of the hoop, its mechanical - energy is the sum of kinetic and potential energy. The potential energy at the bottom is $U = 0$ (taking the bottom as the zero - potential level), and the kinetic energy is $K=\frac{1}{2}mv^{2}$. By conservation of mechanical energy $E_{i}=E_{f}$, so $mgR=\frac{1}{2}mv^{2}$.

Step2: Solve for the velocity at the bottom

From $mgR=\frac{1}{2}mv^{2}$, we can cancel out the mass $m$ on both sides and solve for $v$. We 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 acting on the block is $F_{net}=N - mg$, where $N$ is the normal force and $mg$ is the weight of the block. According to Newton's second law $F_{net}=ma$, and for circular motion at the bottom of the hoop, $a = \frac{v^{2}}{R}$. So $N - mg=m\frac{v^{2}}{R}$.

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

Substitute $v^{2}=2gR$ into $N - mg=m\frac{v^{2}}{R}$. We have $N - mg=m\frac{2gR}{R}$. Simplifying the right - hand side gives $N - mg = 2mg$. Then, add $mg$ to both sides to solve for $N$. We get $N=3mg$.

Answer:

d. $3mg$