a 225 kg crate initially at rest slides down a frictionless ramp that is inclined at 34.3°. if the ramp is…

a 225 kg crate initially at rest slides down a frictionless ramp that is inclined at 34.3°. if the ramp is 20.0 m long, how long does it take for the crate to reach the bottom? a = 5.52 m/s² t = ? s

a 225 kg crate initially at rest slides down a frictionless ramp that is inclined at 34.3°. if the ramp is 20.0 m long, how long does it take for the crate to reach the bottom? a = 5.52 m/s² t = ? s

Answer

Explanation:

Step1: Identify the kinematic - equation

The crate starts from rest ($v_0 = 0$), and the kinematic equation $x=v_0t+\frac{1}{2}at^{2}$ is used. Since $v_0 = 0$, the equation simplifies to $x=\frac{1}{2}at^{2}$.

Step2: Solve for time $t$

We know that $x = 20.0$ m and $a=5.52$ m/s². Rearranging the equation $x=\frac{1}{2}at^{2}$ for $t$, we get $t=\sqrt{\frac{2x}{a}}$. Substitute $x = 20.0$ m and $a = 5.52$ m/s² into the formula: $t=\sqrt{\frac{2\times20.0}{5.52}}=\sqrt{\frac{40.0}{5.52}}\approx\sqrt{7.246}\approx 2.69$ s

Answer:

$2.69$ s