Frictionless Inclined Plane

1
Frictionless Inclined Plane

• Example 2

2
A block of wood is placed on an inclined plane.
It slides down the frictionless surface. The
plane is inclined 10 degrees. Assuming the block
starts from rest, how long does it take to move 1
meter? How fast is it traveling and what is its
total acceleration?
What is the next step?
If you said to make a drawing, you are right! Do
so now.
3
The Drawing
Anything else that needs to be added?
Oops, I forgot the origin and axes again!
1 meter
o
4
The Data
Click when you have written down your x data.
xo 0m (initial x position) x 1m (final x
position) vox 0m/s (initial x velocity)
Click when you have written down your y data.
yo 0m (initial y position) voy
0m/s (initial y velocity)
Note how the labels have been shortened so that
are easier to write.
5
The Equations
Click when you have written down your equations.
x xo voxt 1/2axt2 (eq 1) vx vox
axt (eq 2)
Did you forget your y equations?
y yo voyt 1/2ayt2 (eq 3) vy voy
ayt (eq 4)
6
Are we ready to plug in numbers yet?
Not yet. What has been left out?
What can you say about the x and y acceleration?
The block moves only in the x direction, so the y
acceleration is clearly zero. What about the x
acceleration? Try making a drawing and see if
you can figure it out.
7
The x and y accelerations
ax gsin(q) ay 0m/s
Why is ay zero?
Because the block cannot fall through the
inclined plane! This will be discussed in more
detail in a later chapter.
8
Time to put it together!
Put all of your data into the formulas and click
to compare.
1m 0m 0m/st 1/2gsin(10o)t2 (eq 1) vx
0m/s gsin(10o)t (eq 2)
y 0m 0m/st 1/2 0m/s2t2 (eq 3) vy 0m/s
0m/s2t (eq 4)
Now simplify your equations, removing all zeroes.
9
Are we there yet?
1m 1/2gsin(10o)t2 (eq 1) vx
gsin(10o)t (eq 2)
y 0m (eq 3) vy 0m/s (eq 4)
We now know the final y position and y
velocity. There are two remaining equations and
two unknowns so we are ready to begin the algebra!
10
I wish I knew the right way to do the algebra!
Here is where some common sense comes in
handy. You have two unknowns, vx and t. You can
solve for vx from the second equation but you
cannot plug it into the first since it isnt
found in the first. You can solve for t using
either the first or second equation and plug it
into the remaining equation. Either will
work! However, if you noticed, the first equation
contains only one unknown, namely t. So that one
might be faster to solve for, with less
algebra. Lets do it!
11
The last steps
t A 1.08s
Note that there is always a sign uncertainty when
you take a square root. Clearly our block moves
downward when time is positive, so we choose the
positive sign.
t 1.08s
vx 1.84m/s
12
Does this make sense?
The easiest way to check your result is to put in
the time back into your original equation to see
if you get the expected distance of 1 meter.
x ½9.8m/s2sin(10o)(1.08s)2 (eq 2) 1.0m