We have a block on a 2-meter-long platform. The block begins to slip when the platform is raised to a height of 1.03 meters, and it takes 2.2 seconds to slide down to the other end once the block starts moving. Our task is to find:

1. The coefficient of static friction ($\mu_s$) between the block and the platform.
2. The coefficient of kinetic friction ($\mu_k$).

Step 1: Determine the angle at which the block begins to slip

The block slips when the angle of the platform relative to the ground reaches a critical value. The height of the raised end is 1.03 m, and the length of the platform is 2 m. Using the geometry of a right triangle, we can calculate the angle of inclination $\theta$:

$\sin(\theta) = \frac{\text{height}}{\text{length}} = \frac{1.03}{2}$ $\theta = \sin^{-1}\left(\frac{1.03}{2}\right) \approx \sin^{-1}(0.515) \approx 31^\circ$

Step 2: Calculate the coefficient of static friction ($\mu_s$)

The block begins to slip when the component of the gravitational force along the incline exceeds the maximum static friction force. The condition for slipping is:

$\mu_s = \tan(\theta)$

Where $\theta$ is the angle of the incline. Using $\theta = 31^\circ$:

$\mu_s = \tan(31^\circ) \approx 0.601$

Thus, the coefficient of static friction is approximately $\mu_s \approx 0.601$.

Step 3: Calculate the coefficient of kinetic friction ($\mu_k$)

Once the block starts sliding, we can analyze the motion to find the coefficient of kinetic friction. The block slides 2 meters in 2.2 seconds. First, we'll find the acceleration of the block using the kinematic equation:

$d = \frac{1}{2} a t^2$ Where:

• $d = 2 \, \text{m}$ (the length of the platform),
• $t = 2.2 \, \text{s}$,
• $a$ is the acceleration of the block.

Solving for $a$:

$2 = \frac{1}{2} a (2.2)^2$ $a = \frac{2 \times 2}{(2.2)^2} \approx \frac{4}{4.84} \approx 0.826 \, \text{m/s}^2$

Next, we apply Newton's second law along the incline:

$m g \sin(\theta) - \mu_k m g \cos(\theta) = m a$

Canceling the mass ($m$):

$g \sin(\theta) - \mu_k g \cos(\theta) = a$

Substitute known values ($g = 9.8 \, \text{m/s}^2$, $\sin(31^\circ) \approx 0.515$, $\cos(31^\circ) \approx 0.857$):

$9.8 \times 0.515 - \mu_k \times 9.8 \times 0.857 = 0.826$ $5.047 - 8.398 \mu_k = 0.826$ $8.398 \mu_k = 4.221$ $\mu_k \approx \frac{4.221}{8.398} \approx 0.503$

Thus, the coefficient of kinetic friction is approximately $\mu_k \approx 0.503$.

Final Answers:

• The coefficient of static friction: $\mu_s \approx 0.601$
• The coefficient of kinetic friction: $\mu_k \approx 0.503$

Would you like further details on any step or have any questions?

Here are five related questions:

1. How would the results change if the platform length were longer?
2. What if the height was less than 1.03 m—how would that affect $\mu_s$?
3. How does increasing the mass of the block affect the coefficients of friction?
4. What are the conditions under which kinetic friction becomes greater than static friction?
5. Can we calculate the block's final velocity after sliding 2 m down the platform?

Tip: Always verify which type of friction applies before solving. Static friction acts until motion begins, and kinetic friction takes over once the object is in motion.