# A 5.0kg mass is released from rest at the top of a frictionless 30^@ inclined plane. What is the kinetic energy of the mass when it has slid down 40cm down the incline starting from rest?

Dec 9, 2017

Since it is a frictionless surface the gain in kinetic energy is equal to the loss in potential energy which is 9.8 joules.

#### Explanation:

Since it is a frictionless surface the gain in kinetic energy is equal to the loss in potential energy. We need to first find the vertical height dropped to find the loss in potential energy.

Since the ramp is at a 30° angle, the decrease in height when descending 40 cm down the ramp is found using either the 30, 60, 90° triangle relationships or the trigonometric functions.

Using the 30, 60, 90° triangle relationships the ratios of the length of the triangle's sides are $1 , \sqrt{3} , 2$ opposite respectively the 30, 60, and 90 degree angles.

Thus, since the 40 cm side is opposite the 90 degree angle the side opposite the 30° angle will be 1/2 of that length, or 20 centimeters.

Or using trigonometry we could say the $\sin \left(30\right) = \frac{o p p o s i t e}{h y p o t e \nu s e} = \frac{h e i g h t}{h y p o t e \nu s e} = \frac{h}{40}$
So
$40 \sin \left(30\right) = h$
$40 \cdot .5 = h$
And thus $h = 20 c m$ the same as found above.

The Gain in kinetic energy is the loss in potential energy, $m g h$ where $m$ is the mass, $g$ is the acceleration of gravity, and $h$ is the loss in height.

$m g h$=$5.0 k g \cdot 9.8 \left(\frac{m}{s} ^ 2\right) \cdot .2 m$=$9.8$ joules

So the gain in kinetic energy is 9.8 joules.