A block of ice with mass 2.00 kg slides 1.35 m down an inclined plane that slopes downward at an angle of 36.9° below the horizontal. If the block of ice starts from rest, what is its final speed?

The given data can be listed below,

• The mass of the ice block is, m = 2.00 kg
• The distance slid down by the ice block is, s = 1.35 m
• The angle of inclination is,$\theta =36.9°$

The orientation of the body under examination is referred to as the angle of inclination, whether it be a vertical or a horizontal angle.

According to the work-energy theorem, the work done is given by,

$W_{tot}=\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2$

Here, m is the mass of the ice block,$v_f$ is the final velocity of the ice block and$v_i$ is the initial velocity of the ice block.

Substitute all the values in the above to find the final speed of the ice block as:

$$\begin{gathered} W_{tot}=\frac{1}{2}m{v}_f^2 \\ mgs\sin \theta =\frac{1}{2}m{v}_f^2 \\ {v}_f=\sqrt{2gs\sin \theta } \\ =\sqrt{2\times \left(9.8m/s^2\right)\left(1.35m\right)\sin 36.9°} \\ =\sqrt{26.46\sin 36.9°}m/s \\ =3.99m/s \end{gathered}$$

Thus, the final speed of the ice block is 3.99 m/s.