A 0.360 kg particle slides around a horizontal track. The track has a smooth, ve

Question

A 0.360 kg particle slides around a horizontal track. The track has a smooth, vertical outer wall forming a circle with a radius of 1.27 m. The particle is given an initial speed of 8.73 m/s. After one revolution, its speed has dropped to 6.02 m/s because of friction with the rough floor of the track. Calculate the energy loss due to friction in one revolution.

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Calculate the coefficient of kinetic friction.

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What is the total number of revolutions the particle makes before stopping?

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PLEASE SHOW WORK

Explanation / Answer

energy loss is 0.5*m*(v^2-u^2) = 0.5*0.36*(6.02^2-8.73^2) =-7.19 J

then Work done by the frictional force = loss in energy

fs*S = 7.19

mu_s*m*g*2*3.142*1.27 = 7.19

coefficient of friction is mu_s = 7.19/(2*3.142*1.27*0.36*9.81) =0.255

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using v^2-u^2 = 2*a*S

(6.02^2-8.73^2) = 2*a*2*3.142*1.27

accelaration a = -2.504 m/s^2

then again using v^2-u^2 = 2*a*S

(0^2-8.73^2) = -2*2.504*S

S = 15.2 m = N*2*pi*r

no.of revolutions are N = 15.2/(2*3.142*1.27) = 1.9 rev = 2 rev appoximately

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