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Chapter 3: Q3.4-18E (page 116)

When an object slides on a surface, it encounters a resistance force called friction. This force has a magnitude of μ±· , whereμ the coefficient of kinetic friction and N isthe magnitude of the normal force that the surface applies to the object. Suppose an object of mass30 kgis released from the top of an inclined plane that is inclined 30°to the horizontal (see Figure 3.11). Assume the gravitational force is constant, air resistance is negligible, and the coefficient of kinetic frictionμ=0.2 . Determine the equation of motion for the object as it slides down the plane. If the top surface of the plane is 5 m long, what is the velocity of the object when it reaches the bottom?

Short Answer

Expert verified

The velocity of the object when it reaches the bottom isVt=5.66m/sec and xt=5.66t+c.

Step by step solution

01

Find the value of velocity

There are two forces are written as,

F1=mg sin 30oF2=-μ mg cos 30o

Now put the given values then;

mdvdt=mgsin30o-μmgcos30odvdt=gsin30o-μgcos30o                   Puttingthevaluesof g=0.2dvdt=3.207

Sincevt=Vxt,

So, the values are written as;

dvdt=VdVdxVdVdx=3.207

02

Find the value of velocity by limits.

Now, find the value of velocity then,

∫VdV=∫053.207dxV22-3.207x05=0                                        Integratewithlimits0to5V2=32.07Vt=5.66m/sec

03

Evaluate the equation of motion.

Now, for the value of x(t),

v=5.66dxdt=5.66x t=5.66 t+c

Therefore, the results are xt=5.66 t+c and vt=5.66 m/sec.

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