A particle can slide along a track with elevated ends and a flat central part, a

Question

A particle can slide along a track with elevated ends and a flat central part, as shown below. The flat part has length L = 40 cm. The curved portions of the track are frictionless, but for the flat part the coefficient of kinetic friction is µk = 0.31. The particle is released from rest at point A, which is a height h = L/2. How far from the left edge of the flat does the particle finally stop?

A particle can slide along a track with elevated ends and a flat central part, as shown below. The flat part has length L = 40 cm. The curved portions of the track are frictionless, but for the flat part the coefficient of kinetic friction is Muk = 0.31. The particle is released from rest at point A, which is a height h = L/2. How far from the left edge of the flat does the particle finally stop?

Explanation / Answer

energy of the block after just reaching the friction track

Ei = 1/2* mv^2 = mg h ,

Ei = m* g* 0.2 = 0.2 mg Joule

Now , This energy will get converted to -ve work done by friction.

so ,
Friction = mu * N ,

so,

F = 0.31* m*g N

so ,
maximum distance it can travel ,

x = Ei/ Friction

x= 0.2 /0.31 = 0.645 m = 64.5 m

As it is greater than L=0.4 , so it will reach the end . And then retrace its path , coming back to friction flat plate.

Till now ( at just the start of the travel of backward path on Flat)
energy spend by block = friction* 0.4 = 0.31*0.4*mg = 0.124 mg Joules

so ,
energy remaining = Ei - 0.124 mg = 0.2 mg- 0.124*mg = 0.076 *mg Joules,

This is the energy remaining to be lose by friction ,

so ,

if the block moves x' m in the left direction before it stops ,

x' = 0.076 mg / ( 0.31 mg ) = 0.245 m = 24.5 cm

so ,
Its distance from left edge = 40-24.5 = 15.5 cm

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