A wet bar of soap slides down a ramp 9.0 m long inclined at 8.0°. How long does it take to reach the bottom? Assume${\mu }_k=\mathbf{0}\mathbf{.}\mathbf{060}$.

A wet soap bar slides down the ramp inclined at an8.0°angle.Since the soap slides downward on the ramp, a kinetic frictional force acts on the soap in the direction opposite to the bar’s motion.

Resolve the mass components along the x-axis and y-axisdirections. Then, apply the equilibrium conditions and evaluate the time taken by the soap to reach the bottom of the ramp.

The given data can be listed below as:

• The length of the ramp is $L=9\mathrm{m}$.
• The value of the kinetic friction coefficient is ${\mu }_k=0.060$.
• The angle of inclination of the ramp is $\theta =8°$.
• The acceleration due to gravity is $g=9.81\mathrm{m}/{\mathrm{s}}^2$.
• The initial velocity of the bar is $u=0\mathrm{m}/\mathrm{s}$.

The free body diagram of the soap bar can be shown as:

Here, $F_N$ is the normal reaction force on the soap bar, $F_{fr}$ is the frictional force, g is the acceleration due to gravity, mgis the weight of the bar, and m is the mass of the bar.

The bar is sliding down the ramp with some acceleration (a).

At the equilibrium condition, the forces along the horizontal direction can be expressed as:

$\begin{array}{c}\sum F_x=ma\\ mg\sin \theta -F_{fr}=ma\\ F_{fr}=mg\sin \theta -ma\dots \left(i\right)\end{array}$

At the equilibrium condition, the forces along the vertical direction can be expressed as:

$\begin{array}{c}\sum F_y=0\\ mg\cos \theta -F_N=0\\ F_N=mg\cos \theta \dots \left(ii\right)\end{array}$

The frictional force can be expressed as:

$F_{fr}={\mu }_k{F}_N$

Substitute the values of (i) and (ii) in the above expression.

$\begin{array}{c}mg\sin \theta -ma={\mu }_{k}mg\cos \theta \\ a=g\sin \theta -{\mu }_{k}g\cos \theta \end{array}$

Substitute the values in the above expression.

$\begin{array}{c}a=9.81\mathrm{m}/{\mathrm{s}}^2\times \sin 8°-0.060\times 9.81\mathrm{m}/{\mathrm{s}}^2\times \cos 8°\\ =0.78\mathrm{m}/{\mathrm{s}}^2\end{array}$

The distance moved by the soap can be expressed as:

$L=ut+\frac{1}{2}a{t}^2$

Here, L is the distance moved by the wet soap bar, which is equal to the length of the bar,t is the time taken by the bar to reach the bottom, and $u$is the initial velocity of the bar whose value is 0 m/s.

Substitute the value of the initial velocity of the bar in the above expression.

$\begin{array}{c}L=0\mathrm{m}/{\mathrm{s}}^2\times t+\frac{1}{2}a{t}^2\\ \frac{1}{2}a{t}^2=L-0\\ t^2=\frac{2L}{a}\\ t=\sqrt{\frac{2L}{a}}\end{array}$

Substitute the value of the initial velocity of the bar in the above expression.

$\begin{array}{c}t=\sqrt{\frac{2\times 9\mathrm{m}}{0.78\mathrm{m}/{\mathrm{s}}^2}}\\ =4.80\mathrm{s}\end{array}$

Thus, the time taken by the soap bar to reach the bottom is 4.80 s.