Question

A block of ice with mass 2.00 kg slides 0.750 m down an inclined plane that slopes downward at an angle of 36.9 degrees below the horizontal. If the block of ice starts from rest, what is its final speed? You can ignore friction.

Answer 1

Concepts and reason

The concepts required to solve this problem are Newton’s second law and kinematics equation of motion.

Initially, Draw the free body diagram from given data. Then find the acceleration of block of ice on inclined plane by using Newton’s second law of motion. Finally, calculate the final speed of block by using kinematic equation.

Fundamentals

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

$\sum {F = ma}$

Here, $\sum F$ is the sum of all the forces on the object, $m$ is mass of the object, and $a$ is the acceleration of the object.

The kinematics equation of motion is,

$v_{\rm{f}}^2 = v_{\rm{i}}^2 + 2ad$

Here, $d$ is the distance travelled by the object, ${v_{\rm{i}}}$ is the initial velocity, ${v_{\rm{f}}}$ is the final velocity, and $a$ is the acceleration.

Draw the following free body diagram of block from given data:

Here, m is mass of block, g is acceleration due to gravity, N is normal force on the block, and $\theta$ is angle made by incline with horizontal.

The acceleration a of the block on the incline plane is equal to $g\sin \theta .$

$a = g\sin \theta$

Use the following equation of kinematics to find the final speed of block.

$v_{\rm{f}}^2 = v_{\rm{i}}^2 + 2ad$

Here, ${v_{\rm{f}}}$ is final speed, ${v_{\rm{i}}}$ is initial speed, a is acceleration, and d is distance.

Substitute $g\sin \theta$ for $a$ in the above equation and solve for final speed ${v_{\rm{f}}}$ .

$\begin{array}{c}\\v_{\rm{f}}^2 = v_{\rm{i}}^2 + 2\left( {g\sin \theta } \right)d\\\\{v_{\rm{f}}} = \sqrt {v_{\rm{i}}^2 + 2\left( {g\sin \theta } \right)d} \\\end{array}$

Substitute $0{\rm{ m/s}}$ for ${v_{\rm{i}}}$ , $9.8{\rm{ m/}}{{\rm{s}}^2}$ for $g$ , $36.9^\circ$ for $\theta$ , and $0.750{\rm{ m}}$ for $d$ in the equation ${v_{\rm{f}}} = \sqrt {v_{\rm{i}}^2 + 2\left( {g\sin \theta } \right)d}$ and calculate ${v_{\rm{f}}}$ .

$\begin{array}{c}\\{v_{\rm{f}}} = \sqrt {{{\left( {0{\rm{ m/s}}} \right)}^2} + 2\left( {\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\sin 36.9^\circ } \right)\left( {0.750{\rm{ m}}} \right)} \\\\ = 2.97{\rm{ m/s}}\\\end{array}$

Ans:

The final speed of the block of ice is $2.97{\rm{ m/s}}$ .