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 0.105 m on a horizontal surface. At a point a distance 6.50×102 m away from equilibrium, the speed of the plate is 0.360 m/s . What is the period? What is the displacement when the speed is 0.170 m/s ? 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

Here ,

mass ,m = 0.24 Kg

amplitude , A = 0.105

let the spring constant is k

Using conservation of energy

0.5 * k * 0.105^2 = 0.5 * 0.24 * 0.36^2 + 0.5 * k * (0.065)^2

solving for k

k = 4.57 N/m

time period = 2 * pi * sqrt(m/k)

time period = 2 * pi * sqrt(0.24/4.57)

time period = 1.44 s

the time period is 1.44 s

---------------------------------
let the displacement is d

Using conservation of energy

0.5 * 4.57 * 0.105^2 = 0.5 * 0.24 * 0.17^2 + 0.5 * 4.57 * (x)^2

x = 0.0975 m

the displacement is 0.0975 m

-------------------------------

here , let the coefficient of friction is u

u * g = A * (2pi/T)^2

u * 9.8 = 0.105 * (2*pi/1.44)^2

u = 0.204

the coefficient of static friction is 0.204

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.