Question

Two blocks $m_1$ = 5 kg and $m_2$ = 2 kg are connected at the two ends of a spring of force constant k = 100Nm$^{-1}$. Friction coefficient between $m_1$ and ground is 0.4, and between $m_2$ and ground is 0.2 . The minimum horizontal velocity v that must be imparted to the $m_2$ towards right in order to just move $m_1$ over surface is: (g = 10 m/s$^2$)

• A. $\sqrt{1.4}$ m/s
• B. $\sqrt{1.8}$ m/s
• C. $\sqrt{2.2}$ m/s
• D. $\sqrt{2.8}$ m/s

Answer 1

Answer: (D)

$\sqrt{2.8}$ m/s

Answer 2

To solve this problem, we need to determine the minimum initial velocity v imparted to block m_2 such that block m_1 just begins to move.

Step 1: Determine the critical spring compression x_0 for m_1 to just move.

For m_1 to just move, the force exerted by the spring (F_s) on m_1 must be equal to the maximum static friction force (f_{s1,max}) on m_1. The maximum static friction force on m_1 is given by: $f_{s1,max} = \mu_1 N_1 = \mu_1 m_1 g$

The spring force when compressed by x_0 is: $F_s = k x_0$

Equating these two forces for m_1 to just move: $k x_0 = \mu_1 m_1 g$ $x_0 = \frac{\mu_1 m_1 g}{k}$

Given values: $m_1 = 5 \text{ kg}$ $\mu_1 = 0.4$ $g = 10 \text{ m/s}^2$ $k = 100 \text{ Nm}^{-1}$

Substitute the values: $x_0 = \frac{0.4 \times 5 \times 10}{100} = \frac{20}{100} = 0.2 \text{ m}$

Step 2: Apply the Work-Energy Theorem to the system of m_2 and the spring.

For the minimum initial velocity v for m_2, we assume that m_2 comes to a momentary stop (its velocity becomes zero) exactly when the spring is compressed by x_0. This ensures that all the initial kinetic energy of m_2 is used to compress the spring and overcome the friction on m_2.

Consider the initial state (when m_2 is given velocity v, spring is at natural length) and the final state (when m_2 momentarily stops and the spring is compressed by x_0).

Initial Kinetic Energy of m_2: $KE_i = \frac{1}{2} m_2 v^2$ Final Kinetic Energy of m_2: $KE_f = 0$ (since m_2 momentarily stops)

Initial Spring Potential Energy: $PE_i = 0$ (spring is at natural length) Final Spring Potential Energy: $PE_f = \frac{1}{2} k x_0^2$

Work done by friction on m_2 ($W_{f2}$): Friction force on m_2 is $f_2 = \mu_2 N_2 = \mu_2 m_2 g$. This force opposes the displacement x_0. $W_{f2} = -f_2 x_0 = -\mu_2 m_2 g x_0$

According to the Work-Energy Theorem: $\Delta KE + \Delta PE = W_{f2}$ $(KE_f - KE_i) + (PE_f - PE_i) = W_{f2}$ $(0 - \frac{1}{2} m_2 v^2) + (\frac{1}{2} k x_0^2 - 0) = -\mu_2 m_2 g x_0$ $-\frac{1}{2} m_2 v^2 + \frac{1}{2} k x_0^2 = -\mu_2 m_2 g x_0$

Rearrange the equation to solve for $v^2$: $\frac{1}{2} m_2 v^2 = \frac{1}{2} k x_0^2 + \mu_2 m_2 g x_0$ Multiply by 2: $m_2 v^2 = k x_0^2 + 2 \mu_2 m_2 g x_0$ $v^2 = \frac{k x_0^2}{m_2} + 2 \mu_2 g x_0$

Given values: $m_2 = 2 \text{ kg}$ $\mu_2 = 0.2$ $g = 10 \text{ m/s}^2$ $k = 100 \text{ Nm}^{-1}$ $x_0 = 0.2 \text{ m}$

Substitute these values into the equation for $v^2$: $v^2 = \frac{100 \times (0.2)^2}{2} + 2 \times 0.2 \times 10 \times 0.2$ $v^2 = \frac{100 \times 0.04}{2} + 0.8$ $v^2 = \frac{4}{2} + 0.8$ $v^2 = 2 + 0.8$ $v^2 = 2.8$

Therefore, the minimum horizontal velocity v is: $v = \sqrt{2.8} \text{ m/s}$