91Ó°ÊÓ

The given data can be listed below as,

• The mass of the rock is, $m=1.20kg.$
• The radius of the hemispherical bowl is, $R=0.50m.$
• The work done by friction on the rock when it moves from point A to point B at the bottom of the bowl is, $W_f=0.22J.$

Whenever an object moves along a specific circular path, there would be a force that acts centripetal force on the object. The relation between the centripetal force and the object mass is a direct linear one.

The relation to calculate the workdone on the rock by the normal force is expressed as,

$W_N=Nd\cos \mathrm{θ}$

Here, $W_N$is the workdone on the rock by the normal force, is the normal force, is the displacement of the rock and $\mathrm{θ}$ is the angle between the normal force and displacement of the rock.

Since the normal force is perpendicular to the displacement at every point of motion, then the value of angle $\mathrm{θ}$ would be $\mathrm{θ}={90}^0$.

Substitute all the known values in the above equation.

$$\begin{gathered} W_N=Nd\cos {90}^0 \\ =Nd\left(0\right) \\ =0 \end{gathered}$$

Thus, the workdone on the rock by the normal force is 0.

The relation to calculate the workdone on the rock by the gravity force is expressed as,

$W_g=mgR$

Here, $W_g$is the workdone on the rock by the gravity force, g is the gravitational acceleration whose value is $9.81m/s^2$.

Substitute all the known values in the above equation.

$$\begin{gathered} W_g=\left(0.20kg\right)\left(9.81m/s^2\right)\left(0.50m\right) \\ =0.981kg{m}^2/s^2 \\ =\left(0.981kg{m}^2/s^2\right)×\left(\frac{1J}{1kg.m^2/s^2}\right) \\ =0.981J \end{gathered}$$

Thus, the workdone on the rock by the gravity force is 0.981 J.

The relation to calculate the speed of the rock as its reaches point B is expressed as,

$$\begin{gathered} KE=\left(W_g-W_f\right) \\ \frac{1}{2}m{v}^2=\left(W_g-W_f\right) \\ v=\sqrt{\frac{2\left(W_g-W_f\right)}{m}} \end{gathered}$$

Here, is the kinetic energy of the rock at the bottom B, v is the speed of the rock as its reaches point B.

Substitute all the known values in the above equation.

$$\begin{gathered} v=\sqrt{\frac{2\left(0.981J-0.22J\right)}{0.20kg}\left(\frac{1m^2/s^2}{1J/Kg}\right)} \\ ≈2.76m/s \end{gathered}$$

Thus, the speed of the rock as its reaches point B is 2.76 m/s.

The three forces that act on the rock as it slides down the bowl are gravity force (weight force), normal force, and friction force. The gravity force (weight force) is a constant force that has the same magnitude and direction through the slides of the rock in the bowl, whereas the normal force and friction force are not constant.

Thus, the three forces are gravity force (weight force), normal force, and friction force, and the gravity force is constant, whereas the normal force and friction force are not constant.

The relation to calculate the normal force on the rock due to the bottom of the bowl is expressed as,

$$\begin{gathered} N_1=W_g+F_c \\ =mg+\frac{m{v}^2}{R} \\ =m\left(g+\frac{v^2}{R}\right) \end{gathered}$$

Here, N is the normal force on the rock due to the bottom of the bowl, $F_c$ is the centripetal force and g is the gravitational acceleration whose value is $9.18m/s^2$.

Substitute all the known values in the above equation.

$$\begin{gathered} N_1=\left(0.20kg\right)\left[\left(9.81m/s^2\right)+\frac{{\left(2.76m/s\right)}^2}{\left(0.50m\right)}\right] \\ ≈5.0kgm/s^2 \\ ≈\left(5.0kgm/s^2\right)×\left(\frac{1N}{1kgm/s^2}\right) \\ ≈5.0N \end{gathered}$$

Thus, the normal force on the rock due to the bottom of the bowl is$5.0N$.