The 0.28-kg particle has a speed v = 11.6 m/s as it passes the position shown. T

Question

The 0.28-kg particle has a speed v = 11.6 m/s as it passes the position shown. The coefficient of kinetic friction between the particle and the vertical-plane track is ?k = 0.19. Determine the magnitude of the total force exerted by the track on the particle. What is the change in speed v^(dot) (positive if speeds up, negative if slows down) of the particle?

Please show all work involved! Will give points to the correct answers.

The 0.28-kg particle has a speed v = 11.6 m/s as it passes the position shown. The coefficient of kinetic friction between the particle and the vertical plane track is ?k = 0.19. Determine the magnitude of the total force exerted by the track on the particle. What is the change in speed v^(dot) (positive if speeds up, negative if slows down) of the particle?

Explanation / Answer

normal force on the track by the particle = (mv^2/r)-mg cos theta

=(0.28 11.6^2 /4.8 )- 0.28*9.8*cos 27 = 5.404 N

so, frictional force in the tangential direction = mew*N = 0.19*5.404 = 1.02N

so, the track exerts normal force and frictional force on particle each 90 degrees to each other

magnitude of total force exerted by the track on the particle. = sqrt( 5.04^2 +1.02^2) = 5.14 N

decleration of particle = (force)/mass = (mg sin theta +frictional force)/mass =( (0.28*9.8*0.5 )+1.02)/0.28

= 8. 54m/s^2

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