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A small mass slides down an inclined plane of inclination θ with the horizontal. The coefficient friction is μ=μox where x is the distance through which the mass slides down and μo​ is a positive constant. Then the distance covered by the mass before it stops is


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Solution

Srep1: Given data

  1. The coefficient of friction is μ=μox.
  2. The angle of the inclined plane is θ.

Diagram

A box with an initial speed of 9 m/s is moving up a ramp. The ramp has a  kinetic friction coefficient of 1/2 and an incline of (2 pi )/3 . How

Step2: Frictional force and acceleration

  1. The frictional force is the force that resists the motion of a body over any surface contact with the body.
  2. The frictional force is defined by the form, f=μN, where, N is the normal force exerted on a body and μ is the coefficient of friction.
  3. Acceleration is the time rate of change in velocity. Acceleration is defined as a=dvdt, where, v is the velocity of the body.

Step4: Finding the acceleration

From the figure, it is clear that the normal force exerted on the body is N=mgcosθ.

As we know, the frictional force is f=μN.

So, the equation of the force of the body from the figure is

F=mgsinθ-frictionalforceorF=mgsinθ-μNorF=mgsinθ-μox.mgcosθorma=mgsinθ-μox.mgcosθ(since,F=massm×accelerationa)ora=gsinθ-μox.cosθ........(1)

Step4: Finding the distance

We know the acceleration, a=dvdt

or a=dvdt=dvdx.dxdtora=dvdx.vsince,v=dxdtora=dvdx.v................(2)

Comparing equations 1 and 2 we get,

gsinθ-μox.cosθ=dvdx.vorgsinθ-μox.cosθdx=vdv.............(3)

Integrating equation 3

0xmgsinθ-μox.cosθdx=00vdvorgsinθ.x-μo.cosθ.x220xm=0orgsinθ.xm-μo.cosθ.xm22=0orgμo.cosθ2.xm=gsinθorxm=2gsinθμo.cosθ.=2μo.tanθ.

So, acceleration of the body is 2μotanθ


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