Question

A block of mass 2kg slides on a rough horizontal surface ($\mu_k = 0.2$) with initial speed 10m/ s. Find the stopping distance

Hint:

Notice that the mass $m$ cancels out in the calculation ($a = \mu_k mg / m = \mu_k g$). The stopping distance of a sliding object is independent of its mass, depending only on its initial speed and the coefficient of friction.

Answer 1

Step 1: Understanding the Concept:
When a block slides on a rough surface, kinetic friction exerts a constant force opposing the motion, causing constant deceleration. The stopping distance can be found using kinematics equations or the Work-Energy Theorem. We will use kinematics.

Step 2: Key Formula or Approach:
1. Find the frictional force: $f_k = \mu_k N = \mu_k mg$
2. Find the acceleration (deceleration): $a = \frac{-f_k}{m} = -\mu_k g$
3. Use the kinematic equation relating velocity, acceleration, and distance: $v^2 = u^2 + 2as$

Step 3: Detailed Explanation:
Given values:
Mass, $m = 2$ kg
Coefficient of kinetic friction, $\mu_k = 0.2$
Initial velocity, $u = 10$ m/s
Final velocity, $v = 0$ m/s (since it stops)
Let's assume the acceleration due to gravity $g = 10$ m/s$^2$ for standard calculation (or $9.8$ m/s$^2$, but $10$ is common when not specified to yield clean numbers).
The acceleration is:
\[ a = -\mu_k g = -0.2 \times 10 = -2 \text{ m/s}^2 \]
Now, substitute the knowns into the kinematic equation:
\[ v^2 = u^2 + 2as \]
\[ 0^2 = 10^2 + 2(-2)s \]
\[ 0 = 100 - 4s \]
\[ 4s = 100 \]
\[ s = \frac{100}{4} = 25 \text{ m} \]
(If $g=9.8$ m/s$^2$ was intended, $a = -0.2 \times 9.8 = -1.96$ m/s$^2$, and $s = 100 / (2 \times 1.96) \approx 25.5$ m).

Step 4: Final Answer:
The stopping distance is 25 meters.