91Ó°ÊÓ

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, ${\mathrm{μ}}_{\mathrm{k}}=0.50$

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,$∆E_{th}={\mathrm{μ}}_{k}mgd$

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,

$\mathrm{Δ}{E}_{th}={\mathrm{μ}}_{k}mgd$

Putting, we get,

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

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 ${\mathrm{μ}}_{\mathrm{k}}$of it. Therefore, we get kinetic energy is,

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

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

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

Measured from the lower plateau

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