Question

a 60.0 kg box initially at rest, slides down a frictionless ramp that is inclined at 24.5^{\circ}.

if the ramp is 13.0 m long, how long does it take for the box to reach the bottom?

a = 4.06\text{ m/s}^2

t = ? \text{ s}

Explanation:

⚡ Using what you learned: newton's laws of motion

Step 1: Identify given values

$$
LATEXBLOCK0
$$

Step 2: Use kinematic equation to solve for time

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

Since \( v_i = 0 \):

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

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

Step 3: Calculate the value of \( t \)

$$ t = \sqrt{\frac{2 \times 13.0}{4.06}} $$

$$ t = \sqrt{\frac{26.0}{4.06}} \approx \sqrt{6.4039} \approx 2.53\text{ s} $$

Answer:

2.53