Honors Physics

Conservation of Energy

A 2 kg block slides a distance of 3.0m down a frictionless ramp angled at 30 degrees from the horizontial. The block's final kinetic energy is 35 J. Determine the following:
A) The initial kinetic energy
B) The initial velocity of the block

Step one - Draw a Picture Find the Initial Potential Energy

First find the initial height $sin(\theta)=\frac{H}{Distance \ traveled\ down\ the\ Ramp}$

$sin(30^{\circ})=\frac{H}{3}$
$H=3sin(30^{\circ})$
$H=1.5m$

Now can find the initial potential energy

$PE_{i}=mgH$
$PE_{i} = 2(9.81)(1.5)$
$PE_{i} = 29.4 J$

Starting with the idea of conservation of energy
we can find the initial kinetic energy
Note the final potential energy is 0 J since
the object is at the bottom of the ramp at a
height of 0 m.

$ME_i=ME_f$
$PE_i+KE_i=PE_f+KE_f$
$29.4+KE_i=0+35$
$KE_i=5.6J$

Now we can solve for the initial velocity
of the block from $KE_i$

$KE_i=\frac{1}{2}m(V_i^2)$
$5.6=\frac{1}{2}(2)V_i^2$
$V_i^2=\frac{2(5.6)}{2}$
$V_i=\sqrt{5.6}$
$V_i=2.4\frac{m}{s}$