Q30E - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

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

Step 1: Identification of given data

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, $\theta =36.9°$

Step 2: Concept/Significance of angle of inclination.

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

Step 3: Determination of the final speed of the block of ice when it starts from rest.

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:

 

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

 

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