Block sliding up an inclined plane lab

[TrpnBils]

Homework Statement

A block is given an initial speed of 3m/s up a 22° inclined plane. How far up the plane will it go? How much time elapses before it returns to its starting point?

Homework Equations

D = 1/2at²
a = Δv/t
Vf² = Vi²+ 2a*d

The Attempt at a Solution

The Vy is 1.12m/s. If you assume 9.8m/s² for gravity, then shouldn't you be able to run it through the second equation above to get a time of 0.22 seconds until it stops (half the amount of time til it returns to its original spot)?

As for distance, if you use the third equation there and gravity as your acceleration, shouldn't you be able to solve it that way?

The answers are apparently 1.2m up the plane and 1.6 seconds round-trip but I don't know how...

 

[Yukoel]

Hello TrpnBils,
Again the acceleration due to gravity has to be resolved parallel to the incline .Then your equations work with the right value of a .
regards
Yukoel

 

[TrpnBils]

Like I asked in my other question here tonight, WHY am I multiplying the sine of the angle by the hypotenuse value rather than dividing by it?

In my mind, 9.8 is the Y value because it points downward on the right triangle that the inclined plane forms. It is opposite the 22 degree angle, and the length of the ramp is the hypotenuse of the right triangle. Sin(theta) = opposite/hypotenuse, right? If that's the case, then to find the hypotenuse "part" of the acceleration, why am I multiplying by sin(theta) rather than dividing by it?

 

[Simon Bridge]

If you mean Vy to be vertical (i.e. opposite the direction of gravity) then that's the hard way to do it. Remember, the block is slowing down in the horizontal direction as well.

Try orienting your axis so the x-axis is pointing up the slope, and resolving gravity (i.e. not velocity) into components. Since you don't have friction (?) you can just find the proportion of g that acts against the motion.

 

[Simon Bridge]

g is the hypotenuse.
there is a component pointing down the slope, call is a, and a component perpendicular to the slope, call it b. These three form a right-angled triangle - the angle between g and b is 22 degrees. You need to find a, which is the opposite side.

 

[TrpnBils]

How/why is g the hypotenuse if gravity's acceleration is pointing straight down, and not down the length of the ramp?

 

[Simon Bridge]

Because your axis are aligned with the slope not with gravity.
Look - just humour us and try it that way and see. The way you are trying is needlessly complicated.

 

[TrpnBils]

Because your axis are aligned with the slope not with gravity.
Look - just humour us and try it that way and see. The way you are trying is needlessly complicated.

haha, yeah I can see that it's needlessly complicated...I'd like to humor you and get the right answer using your method, but I can't logically see why it would work that way. In other words, if I came across another problem like this, I'm sure I could do it using what you've told me, but I still don't know WHY it's working that way.

 

[TrpnBils]

Let me give this another go-round here with your method and see if I can get it...

 

[TrpnBils]

So if I rotate it so that the actual inclined surface of the ramp is now my X-axis, I get this...correct?

 

[TrpnBils]

Acceleration in the vertical direction here would then be 3.67m/s^2, correct?

 

[TrpnBils]

Nevermind...got it. I just needed to walk away for a few minutes and stop thinking about it. Thanks everyone!

 

[Simon Bridge]

Not quite - compare:

... this one resolves forces.
The normal force N is the y axis. W is the direction of gravity.
The components of a force (or, in your case, the resulting acceleration) are always smaller than the actual force.
You have to be careful to identify the angle correctly.

(this one includes friction as f_k which, in your case, is zero.)

You can put your axis any way you like and apply the physics, and get the same answer ... it's just that some orientations make the math easier.
If you align your axis with gravity, which feels intuitively "natural" to do, you have to account for the way the ramp pushes on the block ... this would mean resolving the N force horizontally and vertically. The ramp slows the vertical fall due to gravity and pushes horizontally as well. Messy to think about ... easier to think of the ramp as changing the net force to one that just points down the ramp... then do it in 1D.

[edit]Our posts crossed - well done.
Sometimes a change of scene helps.

 

[TrpnBils]

Ok thanks - I think my problem was that I was trying to adjust ALL the variables (acceleration, velocity, etc). Just rotating, adjusting ONE variable, and then using the others as-is seems easier for sure.