Answer 1

The spring constant \(k=100 \mathrm{~N} / \mathrm{m}\) Mass of the first block is \(M=5 \mathrm{~kg}\)

The oscillation of the spring is \(10 \mathrm{~cm} .\)

(a) During the oscillation, where the accelertation becom es maximum there the block slips.

The acceleration of the oscillator becomes maximum at the extreme point. Hence, the block slips at extreme points.

(b) The only force acting on the top block is friction.

Thus, \(\mu_{s} m g=m a_{\max }\)

Therefore, the coefficient of friction is,

\(\mu_{s}=\frac{a_{\max }}{g}=\frac{A \omega^{2}}{g}=\frac{A(\sqrt{k / M})^{2}}{g}\)

\(=\frac{(0.1 \mathrm{~m})(\sqrt{100 \mathrm{~N} / \mathrm{m} / 5 \mathrm{~kg}})^{2}}{9.8 \mathrm{~m} / \mathrm{s}^{2}}=0.2\)

Answer 2

spring constant , k = 100 N/m

block mass , m= 5 Kg

amplitude , A = 10 cm

A = 0.10 m

A) as the acceleration is maximum at the extremes,

the block B is most likely to slip near the extremes.

B)

let the coefficient of friction is u

u * m * g = A * w^2

u * m * g = A * (k/m)

u * 5 * 9.8 = 0.1 * 100/5

sovling for u

u = 0.0408

the coefficient of frictioanl force is 0.0408