A wet bar of soap slides down a ramp 9.2 m long inclined at 8.0∘ .How long does it take to reach the bottom? Assume μk = 0.056.

Express your answer to two significant figures and include the appropriate units.


The time taken by the bar to reach the bottom t=4.886s


Displacement of the bar S=9.2m

Angle of inclination \theta=8.0^{\circ}

Coefficient of friction factor \mu k=0.056

To find:

How long it takes to reach the bottom ‘t’

Step by Step Explanation:


We know that the formula for weight of the soap bar is given as

F_{g}=m g \sin \theta

The frictional force acting on this soap bar is determined by

F_{f}=\mu m g \cos \theta

To determine the constant acceleration of the bar, we derive as

F=m a

Here F=F_{g}-F_{f} and thus

F_{g}-F_{f}=m am g \sin \theta-\mu m g \cos \theta=m a

a=g \sin \theta-\mu g \cos \theta

WhereF_{g}=Force imparted due to weight

F_{f}=Frictional Force

m=Mass of the bar

g=Acceleration due to gravity

a=Acceleration of the bar

\sin \theta and \cos \theta are the angles involved in the system

If the bar starts from the rest

Equations of motion involved in calculating the displacement of the bar is given as

s=\frac{1}{2} a t^{2}, From this

 a t^{2}=2 s

t^{2}=\frac{2 s}{a}

t=\sqrt{\frac{2 s}{a}}

Where s= displacement or length moved by the bar

a=Acceleration of the bar

t=Time taken to reach bottom

Substitute all the known values in the above equation we get

t=\sqrt{\frac{2 \times 9.2}{a}} and we know that

a=g \sin \theta-\mu g \cos \theta

=9.8 \times \sin 8.0-0.056 \times 9.8 \times \cos 8.0



t=\sqrt{\frac{2 \times 9.2}{0.821}}





Thus the time taken by the bar to reach the bottom is t=4.886s

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