Question

A very slippery block of ice slides down a smooth ramp tilted at angle u. The ice is released from rest at vertical height h above the bottom of the ramp. Find an expression for the speed of the ice at the bottom

Short answer

The final velocity of the ice block is $\sqrt{2gh}$

Step-by-step solution

Step 1.  write the given information

The angle of inclination is $u$

The block of ice is initially at rest, therefore, $v_i=0m/s$
The height at which the block was at rest is $h$

Let the final speed of the block is$v_f$

Step 2. To determine the speed of the ice block

Let the slant height of the ramp is $S$

The acceleration experienced by the block is $a=g\sin \left(u\right)$
here, $g$is the acceleration due to gravity

$\sin \left(u\right)=\frac{h}{S}$

Hence, the acceleration is expressed as$a=g\frac{h}{S}$
Now, write the equation of motion,
${v_f}^2={v_i}^2+2aS$
Substitute the known values,
$$\begin{gathered} {v_f}^2=0+2\left(g\frac{h}{S}\right)S \\ {v_f}^2=2gh \\ {v}_f=\sqrt{2gh} \end{gathered}$$
Therefore, the final velocity of the ice block is$\sqrt{2gh}$