Question

A block slides down a frictionless plane having an inclination of θ = 15.0 degrees. The block starts from rest at the top and the length of the incline is 2.00m (a) Draw a free-body diagram of the block. Find (b) the acceleration of the block and (c) its speed when it reaches the bottom of the incline.

Answer 1

Concepts and reason

The concepts used in this problem are Newton’s second law of motion and Newton’s equations of motion. Find the acceleration of the block by find the net force on the block. Then find the speed of block using equations of motion.

Fundamentals

Newton’s second law of motion:

The Newton’s second law of motion states that “the net force on the system is equal to the product of mass and acceleration of the system.”

The expression of net force is:

$F = ma$

Here, $m$ is the mass and $a$ is the acceleration.

Newton’s equations of motion:

The Newton’s equations of motion are the equations which describes the linear motion of an object. These equations can also be converted to rotational motion.

The equations of motion are:

$\begin{array}{l}\\v - u = at\\\\s = ut + \frac{1}{2}a{t^2}\\\\{v^2} - {u^2} = 2as\\\end{array}$

Here, $v$ is the final speed, $u$ is the initial speed, $a$ is the acceleration, $s$ is the distance and $t$ is the time.

(a)

The free body diagram of the block is:

Here, $mg$ is the force of gravity and $N$ is the normal force.

(b)

The net vertical force on the block is zero. There is only horizontal force i.e. $mg\sin \theta$ . So, the net force is:

${F_{net}} = mg\sin \theta$

According to Newton’s second law, the net force on the block is:

${F_{net}} = ma$

$\begin{array}{l}\\ \Rightarrow ma = mg\sin \theta \\\\{\rm{ }}a = g\sin \theta \\\end{array}$

Substitute $9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$ for $g$ and ${15^{\rm{o}}}$ for $\theta$ .

$\begin{array}{l}\\{\rm{ }}a = \left( {9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\sin {15^{\rm{o}}}\\\\{\rm{ }} = {\bf{2}}{\bf{.54}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\\\end{array}$

(c)

The equation of motion:

${v^2} - {u^2} = 2as$

Substitute $2.54\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$ for $a$ , $2\;{\rm{m}}$ for $s$ and $0$ for $u$ .

$\begin{array}{c}\\{v^2} - 0 = 2\left( {2.54\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\left( {2\;{\rm{m}}} \right)\\\\v = \sqrt {10.16\;{{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^{\rm{2}}}} \\\\ = {\bf{3}}{\bf{.19}}\;{\bf{m/s}}\\\end{array}$

Ans: Part a

The free body diagram is:

Part b

The acceleration of the block is 2.54 m/s².

Part c

The speed of the block is 3.19 m/s.