Question

A 708 kg block is pushed on the slope of a 20◦
frictionless inclined plane to give it an initial
speed of 50 cm/s along the slope when the
block is 2.2 m from the bottom of the incline.
The acceleration of gravity is 9.8 m/s
2. What is the speed of the block at the bottom of the plane?
Answer in units of m/s

Answer 1

goodness. Sorry, usually I at least can get myself started. But with this I don't know

Answer 2

we need a few different ingredients here. first, out equations of motion! this will be the key equation:
\[v _{f}^{2}=v _{o}^{2}+2ax\]vf is the final velocity (this is what we want!). vo is the initial velocity (.5m/s in this problem). a is acceleration, and x is the total distance traveled (2.2 m). notice that we can find the velocity at the bottom if we know the initial velocity, the distance and the acceleration.

key to this problem is to CHOOSE THE RIGHT COORDINATE SYSTEM. 'normally' we have y in the vertical and x in the horizontal. BUT here we want x to go 'along the plane' and y to go directly into the plane! we're allowed to choose this different coordinate plan as long as we keep everything consistent.

NOW the hard part: what is the acceleration due to gravity?? well, gravity accelerates in the downward vertical direction, so PART OF THE ACCELERATION WILL GO ALONG THE PLANE, AND ANOTHER PART WILL GO DIRECTLY INTO THE PLANE. the part going directly into the plane will NOT CHANGE THE VELOCITY OF THE BLOCK HERE.

given that the plane is at a 20 degree angle with the horizontal, can you use sine to figure out what component will go along the plane? check in at this point, and we can go further.

Answer 3

So for that you would do 2.2sin20?

Answer 4

wait, no

Answer 5

sorry, I'm drawing a picture now

Answer 6

good! draw a picture. show the block and the plane with the angle. show the direction of acceleration due to gravity. make your x axis go ALONG THE SURFACE OF THE PLANE--NOT along the horizontal. make your y axis go perpendicular to the plane.

Answer 7

then, see WHAT COMPONENT OF ACCELERATION RUNS ALONG THE PLANE. this will come from drawing a triangle and using the angle of 20 degrees.

Answer 8

i meant draw it on paper! the drawing in this interface is too hard, so gimme a second to find a pic from online?

Answer 9

lol, sorry

Answer 10

i lucked out and found the perfect one! take a look. this one uses the variable w. that's for weight! but we'll just imagine that instead the w vector in the diagram stands for acceleration due to gravity.

Answer 11

the key is to look at and try and understand how they
1) set up the x and y axes
2) found the COMPONENT OF W THAT GOES ALONG THE PLANE. they did this with right triangles using trigonometry--the sine function.

Answer 12

Ah, nice, that lines up well

Answer 13

can you see how the acceleration will point straight downwards (vertical), but a ''component'' of it will go along the plane. find the size of this component by multiplying: \[9.8\sin(20)\]

Answer 14

okay, I understand that

Answer 15

because you're trying to figure out the x component, like you said, that runs perp to gravity

Answer 16

great! so, you now know the initial velocity, the distance traveled, and the acceleration (which we found using trigonometry). we can use the equation way above to find the final velocity!

Answer 17

nice, that's awesome and way easier than I was thinking

Answer 18

the x component is USUALLY perpendicular to gravity:

Answer 19

but we have CHANGED THE DIRECTION OF OUR X AND Y AXES TO MATCH THIS PROBLEM--which is totally allowed in physics!

so we have this instead: