To solve this problem, we need to find the shortest distance $d_{\text{min}}$ over which the large box of mass $M$ can stop without the small box of mass $m$ slipping, given that the initial velocity of the large box is $v_0$ and the coefficient of static friction between the two boxes is $\mu_s$.

Key points:

• For the small box to avoid slipping, the force due to static friction must be enough to prevent the relative motion between the two boxes.
• The static friction force provides the maximum possible horizontal force to accelerate (or decelerate) the small box with the large box.
• We aim to decelerate the large box as quickly as possible without causing the small box to slip.

Step 1: Maximum static friction force

The maximum static friction force $f_{\text{max}}$ that can act between the large and small box is:

$f_{\text{max}} = \mu_s m g$

where:

• $\mu_s$ is the coefficient of static friction,
• $m$ is the mass of the small box,
• $g$ is the acceleration due to gravity.

This static friction force must provide the necessary deceleration to the small box.

Step 2: Maximum acceleration without slipping

The friction force $f_{\text{max}}$ will cause the small box to decelerate along with the large box. The maximum deceleration $a_{\text{max}}$ the large box can have without the small box slipping is:

$f_{\text{max}} = m a_{\text{max}} \quad \Rightarrow \quad a_{\text{max}} = \frac{f_{\text{max}}}{m} = \mu_s g$

Thus, the maximum deceleration of the large box is $a = \mu_s g$.

Step 3: Shortest stopping distance

Now, we can find the shortest stopping distance $d_{\text{min}}$ using kinematic equations. The large box starts with an initial velocity $v_0$ and decelerates at the maximum rate $a = \mu_s g$ until it comes to rest. Using the kinematic equation:

$v_f^2 = v_0^2 + 2 a d$

where:

• $v_f = 0$ is the final velocity (the box stops),
• $v_0$ is the initial velocity,
• $a = -\mu_s g$ is the acceleration (negative because it's deceleration),
• $d_{\text{min}}$ is the stopping distance.

Solving for $d_{\text{min}}$:

$0 = v_0^2 + 2(-\mu_s g) d_{\text{min}} \quad \Rightarrow \quad d_{\text{min}} = \frac{v_0^2}{2 \mu_s g}$

Final Answer:

The shortest distance in which the large box can stop without the small box slipping is:

$d_{\text{min}} = \frac{v_0^2}{2 \mu_s g}$

Do you need more details or have any questions?

Here are some related questions to expand on this topic:

1. What happens if the coefficient of static friction is zero?
2. How does the mass of the large box affect the stopping distance?
3. What is the stopping distance if the initial speed is doubled?
4. How would the presence of kinetic friction affect the solution if slipping occurs?
5. How can the stopping time be calculated from the given parameters?

Tip: In friction problems, always distinguish between static and kinetic friction as they govern different phases of motion.