Question

What minimum speed does a 100 g puck need to make it to the top of a 2.9-m-long, 34° frictionless ramp?

Answer 1

Concept and reason

The concept required to solve this problem is the law of conservation of energy.

The total energy of the puck at the bottom of the ramp is its kinetic energy. If this total kinetic energy is converted into total potential energy of the puck when it reaches the top of the ramp. So, first write the expression for the kinetic energy of the puck when it is at bottom of the ramp and equal this expression to the expression for potential energy at top of the ramp. Now solve for the minimum speed of the puck needed to go to the top of the ramp.

Fundamentals

The expression for kinetic energy K of a moving object of mass m is,

$K = \frac{1}{2}m{v^2}$

Here, v is the speed of the object.

The potential energy of an object which is at a height h above the reference level is,

$U = mgh$

Here, g is the acceleration due to gravity.

A use full trigonometry ratio in a right angle triangle is,

$\sin \theta = \frac{{{\rm{opposite side}}}}{{{\rm{hypotonuse}}}}$

The diagram to calculate height of the ramp using the length of the ramp and inclination of the ramp is as follows:

A use full trigonometry ratio in a right angle triangle is,

$\begin{array}{c}\\\sin \theta = \frac{{{\rm{opposite side}}}}{{{\rm{hypotonuse}}}}\\\\\sin 34^\circ = \frac{h}{{2.9{\rm{ m}}}}\\\end{array}$

Solve for h.

$\begin{array}{c}\\h = \left( {2.9{\rm{ m}}} \right)\sin 34^\circ \\\\ = {\rm{1}}{\rm{.62 m}}\\\end{array}$

The expression for kinetic energy K of a moving object of mass m is,

$K = \frac{1}{2}m{v^2}$

The potential energy of an object which is at a height h above the reference level is,

$U = mgh$

The total energy of the puck at the bottom of the ramp is its kinetic energy. If this total kinetic energy is converted into total potential energy of the puck when it reaches the top of the ramp.

$K = U$

Substitute $\frac{1}{2}m{v^2}$ for K and $mgh$ for U.

$\begin{array}{c}\\\frac{1}{2}m{v^2} = mgh\\\\\frac{1}{2}{v^2} = gh\\\end{array}$

Rearrange the equation for v.

$v = \sqrt {2gh}$

Substitute $9.8{\rm{ m/}}{{\rm{s}}^2}$ for g and 1.62 m for h.

$\begin{array}{c}\\v = \sqrt {2\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {1.62{\rm{ m}}} \right)} \\\\ = {\rm{5}}{\rm{.64 m/s}}\\\end{array}$

Ans:

The required speed of the puck is 5.64 m/s.