Question

The two blocks in oscillate on a frictionless surface with a period of 1.5 s. The upper block just begins to slip when the amplitude is increased to 40 cm. What is the coefficient of static friction between the two blocks?

Answer 1

Answer:

0.72

Explanation:

$T$ = Time period of oscillation = 1.5 s

Angular frequency is given as

$w = \frac{2\pi }{T}\\w = \frac{2(3.14) }{1.5}\\w = 4.2 rad/s$

$A$ = Amplitude of oscillation = 40 cm = 0.40 m

$\mu$ = Coefficient of static friction = ?

$a$ = acceleration of the block

$m$ = mass of the block

Maximum acceleration of the block is given as

$a = Aw^{2}$

frictional force is given as

$f = \mu mg$

As per newton's second law

$f = ma \\\\\mu mg = ma \\\mu g = a\\\mu g = Aw^{2}\\\mu (9.8) = (0.40)(4.2)^{2}\\\mu = 0.72$

Answer 2

The coefficient of static friction between the two blocks is 0.72.

How to calculate the coefficient?

From the information given, the angular frequency will be:

= 2π/T

= 2(3.14)/1.5

= 4.2rad/s

The maximum acceleration of the block is Aw² and the frictional force is umg. In this case,

umg = ma

u(9.8) = 0.40 × 4.2².

u = (0.40 × 4.2²) / 9.8

u = 0.72

Therefore, the coefficient of static friction between the two blocks is 0.72.

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