The figure shows mass m₁ resting on top of mass m₂ which is attached to a spring with spring constant 10 N/m,where m₁ = m₂ = 20 g.The coefficient of static friction between the two masses is 0.6.The masses are oscillating with SMH on a frictionless surface.What is the maximum amplitude of oscillation without m₁ slipping off m₂? A)0.012 m B)0.024 m C)0.036 m D)0.048 m E)0.056 m

Question

Quizplus · Accepted Answer

The maximum amplitude of oscillation without m₁ slipping off m₂ is when the static friction force is at its maximum value, which is given by $f_s = \mu_s mg$, where $\mu_s$ is the coefficient of static friction, $m$ is the mass of the upper block, and $g$ is the acceleration due to gravity. The maximum static friction force is therefore $f_s = (0.6)(0.02 ext{ kg})(9.8 ext{ m/s}^2) = 0.1176 ext{ N}$. At the maximum amplitude of oscillation, the spring is stretched/compressed by the distance $d$ from its equilibrium position, so the force on the upper block is $fs$ in the direction of motion. The force of gravity on the upper block is $mg$ downward, and the force from the spring is $-kx$ upward (where $k$ is the spring constant and $x$ is the displacement from the equilibrium position, which is equivalent to $d$). Therefore, we have the following equation of motion for the upper block: $$fs - mg - kx = 0$$ At the maximum amplitude of oscillation, the velocity is momentarily zero and the acceleration is a maximum, so $fs$ is equal to the maximum static friction force as calculated above. Rearranging the equation, we find: $$d = x = (fs - mg)/k = (0.1176 ext{ N} - (0.02 ext{ kg})(9.8 ext{ m/s}^2))/10 ext{ N/m} = 0.024 ext{ m}$$ Therefore, the maximum amplitude of oscillation without the upper block slipping off is 0.024 m, which corresponds to choice (B).

The Figure Shows Mass M1 Resting on Top of Mass

The Figure Shows Mass M₁ Resting on Top of Mass