two blocks (m = 1.5 kg and M = 8.5 kg) and a spring (k = 240 N/m) are arranged o

Question

two blocks (m = 1.5 kg and M = 8.5 kg) and a spring (k = 240 N/m) are arranged on a horizontal, frictionless surface. The coefficient of static friction between the two blocks is 0.40. What amplitude of simple harmonic motion of the spring–blocks system puts the smaller block on the verge of slipping over the larger block?

In arranged on a horizontal, frictionless surface. The coefficient of a (k 240 N/m) are arranged on a horizontal, frictionless surface. The coefficient of static friction a spring the two harmonic motion of the spring-blocks system puts the smaller block on the verge of slipping over the larger block?

Explanation / Answer

To put the smallerblock on the verge of slipping over the larger block the force mustapply on the smaller block is

F = mg

but F=ma

ma=mg

acceleration a=g

is the coefficient of static friction = 0.4

g is free fall acceleration = 9.8m/s2

But in SHM acceleration a =2xmax

is the angularfrequency of SHM

xmax is the amplitude of SHM

but 2=k/(M+m)

k is spring constant = 240N/m

M=8.5 kg, m=1.5 kg

2=(240N/m)/(8.5kg+1.5kg)

= 24 (rad/s)2

xmax = a/2 =(0.4*9.8m/s2)/(24(rad/s)2)

= 0.1633m

= 16.33cm

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