91Ó°ÊÓ

• The mass of the lunch box, m = 0.500 kg .
• The angle, $\mathrm{θ}=30°$.
• The distance of the lunch box from the curve’s center of curvature, R = 50 m .

The force that acts between two objects in contact is known as friction force. If the object is moving, then there will be a kinetic friction force, and if the object is at rest, then there will be a static friction force.

Draw the free-body diagram of the lunch box.

The force acting on the lunch box along the x-axis is given by,

$$\begin{gathered} T\sin \mathrm{θ}=\underset{}{∑F_x} \\ T\sin \mathrm{θ}=m{a}_{rad} \\ T\sin \mathrm{θ}=\underset{}{{\displaystyle \frac{m{v}^2}{R}}}.......\left(1\right) \end{gathered}$$

Here, T is the tension force, and v is velocity.

The force acting on the lunch box along the y-axis is given by,

$$\begin{gathered} \underset{}{∑}{F}_y=0 \\ T-\cos \mathrm{θ}-mg=0 \\ T\cos \mathrm{θ}=mg..........\left(2\right) \end{gathered}$$

Divide equation (1) by (2) as:

$$\begin{gathered} \tan \mathrm{θ}=\frac{v^2}{gR} \\ {v}^2=gR\tan \mathrm{θ} \\ v=\sqrt{gR\tan \mathrm{θ}}........\left(3\right) \end{gathered}$$

Substitute $9.8m/s^2$ for g , 50 m for R , and localid="1663771589050" $30°$for localid="1663771597304" $\mathrm{θ}$in equation (3), and we get,

$$\begin{gathered} v=\sqrt{\left(9.8m/s^2\right)×\left(50m\right)×\left(\tan \left(30°\right)\right)} \\ =16.82m/s \end{gathered}$$

Therefore, the speed of the bus is 16.82 m/s .