Question

a (0.298\text{ kg}) ball initially at rest, slides down a frictionless ramp that is inclined at (28.5^{circ}). if the ramp is (1.13\text{ m}) long, how long does it take for the ball to reach the bottom?

(a = 4.68\text{ m/s}^2)

(t = ?\text{ s})

Explanation:

🆕 New Concept Discovered: Kinematics: Motion with Constant Acceleration
Using equations to relate distance, acceleration, and time

Step 1: Identify the given values

We are given the following values from the problem statement and the image:

• Initial velocity, \( v_0 = 0 \text{ m/s} \) (since the ball starts at rest)
• Distance along the ramp, \( d = 1.13 \text{ m} \)
• Acceleration down the ramp, \( a = 4.68 \text{ m/s}^2 \)

Step 2: Choose the appropriate kinematic equation

To find the time \( t \) it takes to travel a distance \( d \) under constant acceleration \( a \) starting from rest, we use the kinematic formula:

$$ d = v_0 t + \frac{1}{2} a t^2 $$

Since \( v_0 = 0 \), the equation simplifies to:

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

Step 3: Solve for time \( t \)

Rearrange the formula to isolate \( t \):

$$ t^2 = \frac{2d}{a} $$

$$ t = \sqrt{\frac{2d}{a}} $$

Substitute the given values into the equation:

$$ t = \sqrt{\frac{2 \times 1.13}{4.68}} $$

$$ t = \sqrt{\frac{2.26}{4.68}} $$

$$ t = \sqrt{0.4829} $$

$$ t \approx 0.695 \text{ s} $$

Answer:

0.70