91Ó°ÊÓ

Height of the frictionless ramp$\mathrm{h}=3 \mathrm{cm}$

The problem deals with the centripetal acceleration. This is a acceleration of a body traversing a circular path. From the height of the ramp and maximum height, the first hill which the block cannot cross can be found. Using the formula of centripetal acceleration, the hilltop on which the centripetal acceleration of the block is the greatest will be calculated.

Formula:

The centripetal acceleration is given by,

$$\begin{gathered} {\mathrm{a}}_{\mathrm{C}}={\mathrm{v}}^2/\mathrm{r} \\ \mathrm{KE}+\mathrm{PE}=\mathrm{constant} \end{gathered}$$

The height of the frictionless ramp is given as$\text{³ó=3 c³¾}$.

Maximum height that the block should cross is$\mathrm{H}=3.5\mathrm{cm}$.

As$\mathrm{H}>\mathrm{h}$, block cannot cross the fourth hill.

Therefore, the block cannot cross the fourth hill.

As$\mathrm{H}>\mathrm{h}$ , the block cannotcross the fourthhill. Therefore, the block fails to cross the fourth hill.

Thus, it travels back and forth across the path over the first three hills.

It is known that

$\mathrm{KE}+\mathrm{PE}=\mathrm{constant}$

According to the principle of conservation of energy, the gravitational potential energy will be less when kinetic energy is more. Kinetic energy will be more when its speed is more. As the height of the first hill is small, the centripetal acceleration${\mathrm{a}}_{\mathrm{C}}={\mathrm{v}}^2/\mathrm{r}$ will be more.

For the block on the first hilltop

${\mathrm{F}}_{\mathrm{n}}=\mathrm{mg}-\frac{{\mathrm{mv}}^2}{\mathrm{r}}$

Where,

$$\begin{gathered} {\mathrm{F}}_{\mathrm{n}}=\mathrm{normal}\mathrm{force}\mathrm{of}\mathrm{the}\mathrm{block} \\ \mathrm{mg}=\mathrm{gravitational}\mathrm{force}\mathrm{on}\mathrm{the}\mathrm{block} \end{gathered}$$

$\frac{{\mathrm{mv}}^2}{\mathrm{r}}=\mathrm{centripetal}\mathrm{force}$

As the velocity of the block is maximum on the first hilltop, the normal force is minimum on the first hill.