A child with poor table manners is sliding his 240 g dinner plate back and forth

Question

A child with poor table manners is sliding his 240 g dinner plate back and forth in SHM with an amplitude of 0.100 m on a horizontal surface. At a point a distance 6.00×102 m away from equilibrium, the speed of the plate is 0.440 m/s . A)What is the period? B) What is the displacement when the speed is 0.170 m/s ? C) In the center of the dinner plate is a carrot slice of mass 11.0 g . If the carrot slice is just on the verge of slipping at the end point of the path, what is the coefficient of static friction between the carrot slice and the plate?

Explanation / Answer

a)   By law of conservation of energy,

KEmax = Etotal

1/2kA^2 = KEi + PEi

1/2kA^2 = ½kxi^2 + ½mvi^2 ----------------(1)

1/2*k*0.1^2 = ½*k*0.06^2 + ½*0.240*0.440^2

Simplifying k= 7.26 N/m

T = 2*sqrt(m/k) = 2*3.14*sqrt(7.26/0.240) = 34.54s

b)

From(1),

1/2*7.26*0.1^2 = ½*7.26*xf^2 + ½*0.240*0.170^2

Simplifying xf = 0.095 m = 9.5cm

c)

In the center plate has the equilibrium point hence

F= ma

s*mg = kA

s = kA /mg = (7.26*0.1)/(0.011*9.8) = 6.7

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