In Fig.8.57, a block is released from rest at height d =40 cmand slides down a frictionless ramp and onto a first plateau, which has lengthand where the coefficient of kinetic friction is 0.50. If the block is still moving, it then slides down a second frictionless ramp through height d/2and onto a lower plateau, which has length d/2and where the coefficient of kinetic friction is again0.50. If the block is still moving, it then slides up a frictionless ramp until it (momentarily) stops. Where does the block stop? If its final stop is on a plateau, state which one and give the distance Lfrom the left edge of that plateau. If the block reaches the ramp, give the height Habove the lower plateau where it momentarily stops.

A block is released from rest at height

$$\begin{gathered} d=40\mathrm{cm} \\ =0.40\mathrm{m} \end{gathered}$$

The coefficient of kinetic friction is, ${\mathrm{\mu }}_{\mathrm{k}}=0.50$

According to the conservation of energy and using thermal energy generated $\mathbf{∆}{\mathit{E}}_{\mathbf{t}\mathbf{h}}$,we can find the kinetic energy at the end of the first plateau. Considering all forces, we can find the kinetic energy Kas the block starts clamping up. By using this value ofK, we can find the height Habove the lower plateau where it momentarily stops.

Formula:

According to the conservation of energy,

$$\begin{gathered} K=\frac{1}{2}m{v}^2 \\ =mgd \end{gathered}$$

The thermal energy generated is given by,$∆E_{th}={\mu }_{k}mgd$

The gravitational potential energy U is,$U=mgy$

According to the conservation of energy,

$$\begin{gathered} K=\frac{1}{2}m{v}^2 \\ =mgd \end{gathered}$$

But, the thermal energy generated is given by,

$\Delta E_{th}={\mu }_{k}mgd$

Putting, we get,

$$\begin{gathered} K=mgd-{\mu }_{k}mgd \\ =\frac{1}{2}mgd \end{gathered}$$

In its descent to the lowest plateau, it gains $\raisebox{1ex}{$mgd$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$more kinetic energy, but it slides across it loses ${\mathrm{\mu }}_{\mathrm{k}}$of it. Therefore, we get kinetic energy is,

$$\begin{gathered} K=\frac{1}{2}mgd+\frac{1}{2}mgd-\frac{1}{2}{\mu }_{k}mgd \\ K=\frac{3}{4}mgd \end{gathered}$$

Compensating with gravitational potential energyUis,

U =mgy

We get,

$H=\frac{3}{4}d$

Thus, the block stops on the inclined ramp at right, at a height of

$$\begin{gathered} H=0.75d \\ =0.75\times 40 \\ =30\mathrm{cm} \end{gathered}$$

Measured from the lower plateau

Hence, the height H above the lower plateau where it momentarily stops is,H =30 cm