1) A 4kg mass rests on a tabletop and is attached to a spring with the spring co

Question

1) A 4kg mass rests on a tabletop and is attached to a spring with the spring constant 100N/m. The other end of the spring is clamped to the table such that the spring is in horizontal position. A 6.0 kg mass is placed on top of the 4 kg mass. What should be the minimum value of the coefficient of the static friction between the two masses, if an oscillation, whose maximum velocity is 4.0m/s keeps the 6.0 kg mass on top of the 4.0 kg mass?

2) Given the above velocity, find the equation of motion for the oscillator.

Explanation / Answer

1)

m = mas of sysytem = 6 + 4 = 10 kg

w of oscillation = sqrt ( k / m ) = sqrt ( 100 / 10 ) = 3.1623

max velocity of harmonic oscillation = A*w , where A = amplitude

so, A * 3.1623 = 4

so, A = 1.265 m

max acceleration of harmonic syste = A * w^2 = 1.265 * 3.1623^2 = 12.6502 m/sec2

for blocks to stay together ,, the friction should be able to hold the 6 kg block on maximum acceleration

so, 6* 9.8 *( mu) = 6 * 12.6502

so, mu min = 1.2908346

2 )

equation of motuion :

x = 1.265 sin ( 3.1623 * t )

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