Question

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

if the ramp is (13.0\text{ 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: graphical analysis of motion and acceleration

Step 1: Identify given values and kinematic formula

The box starts from rest, meaning its initial velocity is:

$$ v_0 = 0\text{ m/s} $$

The acceleration down the ramp is given as:

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

The distance to travel along the ramp is:

$$ d = 13.0\text{ m} $$

We use the kinematic equation relating distance, initial velocity, acceleration, and time:

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

Step 2: Solve for time \( t \)

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

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

Rearranging the formula to solve for \( t \):

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

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

Substitute the given values:

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

$$ t = \sqrt{\frac{26.0}{4.06}} $$

$$ t = \sqrt{6.4039} $$

$$ t \approx 2.53\text{ s} $$

Answer:

2.53