7a. A 0.402 kg particle slides around a horizontal track. The track has a smooth

Question

7a. A 0.402 kg particle slides around a horizontal track. The track has a smooth, vertical outer wall forming a circle with a radius of 1.37 m. The particle is given an initial speed of 8.19 m/s. After one revolution, its speed has dropped to 6.00 m/s because of friction with the rough floor of the track. Calculate the energy loss due to friction in one revolution.

7b. Calculate the coefficient of kinetic friction.

7c. What is the total number of revolutions the particle makes before stopping? Do not enter units.

Explanation / Answer

apply Kinetic energy KE = 1/2 m(vi^2 - vf^2)

KE = 0.5 * 0.402 * (8.19^2 - 6^2 )

KE lost = 6.24 Joules

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Distance travelled = Circumference C = 2piR

C = 2*3.14 * 1.37

C = 8.6036 m

Work done by friction = Ff * C   

so Ff = 6.24/8.603

Ff = 0.725 N (the frictiona lforce)

Ff = m g

= 0.725 / (0.402* 9.81)

u = 0.183

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The total work done in stopping the particle equals theinitial KE

The initial KE = 1/2 m vi^2

KE = 0.5 * 0.402 * 8.19*8.19

KE = 13.48 J

Then n Ff C = 13.48

so n = 13.48 / (0.725 * 8.603)

n = 2.16 rev

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