Pushing a block up a slope--how far does it go?

[Eclair_de_XII]

Homework Statement

"A block is projected up a frictionless inclined plane with initial speed v₀ = 3.5 m/s. The angle of incline is Φ = 32°. (a) How far up the plane does the block go? (b) How long does it take to get there? (c) What is its speed when it gets back to the bottom?"

Homework Equations

$v=v_0+at$
Answer as given by book: (a) 1.18 m; (b) 0.674 s; (c) 3.5 m/s

The Attempt at a Solution

$\vec v_0=(2.97i+1.85j)\frac{m}{s}$

Now, I figure there's downward acceleration from the normal force and gravity. I set the mass of the block (which originally was not specified) to 1 kg, to simplify my calculations.

$F_N=(sin238°)(1kg)(9.8\frac{m}{s^2})=-8.31N$
$F_g=(cos238°)(1kg)(9.8\frac{m}{s^2})=-5.19N$

So, given this, I find the acceleration for a mass that I invented for the block.

$\vec a = (-8.31i-5.19j)\frac{m}{s^2}$

Then I reduce the initial velocity to zero to find the time, which does not correspond with the answer given by the book.

$-1.85\frac{m}{s}=-9.8\frac{m}{s^2}t$
$t=0.189 s$

As you can see, this number does not match up with the answer as given by the book. Also, I just realized that it should be 32 degrees, and not 36 degrees in the FBD I drew on Paint.

 

[haruspex]

You are inconsistent over your coordinates. You started with i horizontal and j vertical, but in your acceleration equation you seem to have i at the normal and j up the slope. Also, the normal force is not the only force normal to the slope.
It might help if you define your Fg explicitly.

 

[Eclair_de_XII]

$F_g=(9.8\frac{m}{s^2})(cos238°)=-5.19\frac{m}{s^2}$

I don't understand what you mean by "at the normal", "up the slope", and the fact that $F_N$ isn't the only force normal to the slope.

 

[haruspex]

$F_g=(9.8\frac{m}{s^2})(cos238°)=-5.19\frac{m}{s^2}$

I don't understand what you mean by "at the normal", "up the slope", and the fact that $F_N$ isn't the only force normal to the slope.

Define your i and j directions.
What forces act on the block, and in which directions?
Define your "Fg" force - what is it?

 

[Eclair_de_XII]

I don't understand what you mean by "defining my directions." I guess, i and j are the axes relative to the block...?

$F_N$ acts perpendicular to the block, relative from the x- and y-axes I've drawn for it. $F_g$ is the deceleration of the block as it goes up the slope. But because it's at an angle, and not in free-fall, it's not as much as $-9.8\frac{m}{s^2}$.

 

[cnh1995]

Normal force is not needed here, since there's no friction between block and the incline. The block is decelerating due to component of its weight "parallel" to the incline. How will you write that component?

 

[Eclair_de_XII]

On a regular coordinate system, or on the coordinate system relative to the block? If the former, then it's 212°. If the latter, it's just 180°.

 

[haruspex]

I don't understand what you mean by "defining my directions."

I don't understand how I can put it any more plainly. I wrote:

You started with i horizontal and j vertical, but in your acceleration equation you seem to have i at the normal and j up the slope.

(I probably meant "j at the normal and i up the slope")
Which is it? Is i horizontal or is it parallel to the slope?

 

[Eclair_de_XII]

I really don't know. I have no one to teach me this. If I had to take a guess... parallel to the slope.

 

[haruspex]

I really don't know. I have no one to teach me this. If I had to take a guess... parallel to the slope.

It's your co-ordinate system , your choice. You can pick either as long as you are consistent (which you were not).
In the present problem, I would recommend taking i as parallel to the slope and j as normal to it. In those co-ordinates, what are the force components in the i direction?

 

[Eclair_de_XII]

Okay, so it's parallel to the slope, so the angle is 180°...

$F=(cos180°)(1kg)(9.8\frac{m}{s^2})=-9.8\frac{m}{s^2}$

Please forgive me if I'm being dumb. I am literally terrible at physics which is why I dropped the class, and which is why I'm trying to self-study it in preparation to study it in the future.

 

[cnh1995]

Okay, so it's parallel to the slope, so the angle is 180°...

$F=(cos180°)(1kg)(9.8\frac{m}{s^2})=-9.8\frac{m}{s^2}$

Please forgive me if I'm being dumb. I am literally terrible at physics which is why I dropped the class, and which is why I'm trying to self-study it in preparation to study it in the future.

Angle of incline is 32° "w.r.t horizontal". Assume horizontal direction as i and vertical direction as j such that the incline makes 32° with i direction. If you use a little geometry, you'll be able to write the equation for component of weight parallel to the incline in terms of 32° angle.

 

[haruspex]

Angle of incline is 32° "w.r.t horizontal". Assume horizontal direction as i and vertical direction as j such that the incline makes 32° with i direction. If you use a little geometry, you'll be able to write the equation for component of weight parallel to the incline in terms of 32° angle.

Between us we're going to confuse the hell out of Eclair. Under my guidance, i has been chosen as parallel to the slope.

 

[haruspex]

Okay, so it's parallel to the slope, so the angle is 180°...

The angle between what and what is 180?

 

[Eclair_de_XII]

The front and the back?

Should I make it the angle between gravitational force and the incline?

 

[cnh1995]

Between us we're going to confuse the hell out of Eclair. Under my guidance, i has been chosen as parallel to the slope.

Ok. I found #11 confusing, so I tried the other approach. Apologies..

 

[haruspex]

Should I make it the angle between gravitational force and the incline?

Since you want to find the component of gravity down the slope, that would seem to be good move.
Do you understand about finding components? Section 4 of https://www.physicsforums.com/insights/frequently-made-errors-vectors-elementary-use/ might help.

 

[Eclair_de_XII]

Well, I already know the angle between gravity and the slope, thanks to the picture. It's 58°.

I still get the same $-5.19\frac{m}{s^2}$, though.

 

[haruspex]

Well, I already know the angle between gravity and the slope, thanks to the picture. It's 58°.

I still get the same $-5.19\frac{m}{s^2}$, though.

Yes, and that much is absolutely fine. So now you have the acceleration parallel to the slope.
Are you familiar with the SUVAT equations.

 

[Eclair_de_XII]

Oh, for some reason, when I divided the x-component of v₀ by that acceleration, it gave me the wrong number. Yet when I just divide 3.5 by 5.19, I get the correct time. But wait, 3.5 m/s is in the horizontal direction, and shouldn't it not be affected by gravity, even if the latter is decreased somewhat by the slope?

 

[Eclair_de_XII]

Or wait... maybe the gravity has no effect on horizontal motion only if it's perpendicular to the object?

 

[haruspex]

Oh, for some reason, when I divided the x-component of v₀ by that acceleration, it gave me the wrong number. Yet when I just divide 3.5 by 5.19, I get the correct time. But wait, 3.5 m/s is in the horizontal direction, and shouldn't it not be affected by gravity, even if the latter is decreased somewhat by the slope?

Your diagram showed it as a horizontal speed, but the problem states

A block is projected up a frictionless inclined plane with initial speed...

 

[Eclair_de_XII]

Yeah, I think drawing pictures helps me analyze this better. Here's the picture for the normal axes, relative to the body:

And here's how I visualize (cos 238°)(g) to negate v₀...

Oh, sorry; in the first picture, -112° should be -122°.

 

[haruspex]

@Eclair_de_XII , I just noticed this was marked solved, but do not see any post that supports that. Did you mark it solved?
Anyway, I've put it back to unsolved. If you have solved it, please post to that effect (preferably with solution).

 

[Eclair_de_XII]

Oh, sorry about that. I just did that on my own, and figured I had this solved. I never really thought I had to provide evidence of this being solved. Anyway:

$v=v_0+at$
$0\frac{m}{s}=3.5\frac{m}{s}-(5.19\frac{m}{s^2})(t)$
$-3.5\frac{m}{s}=-(5.19\frac{m}{s^2})(t)$
$t=0.674s$

$x=(3.5\frac{m}{s})(0.674s)-\frac{1}{2}(5.19\frac{m}{s^2})(0.674s)^2=2.36 m - 1.18m = 1.179m$

$-1.179m=v(0.674s)-\frac{1}{2}(-5.19\frac{m}{s^2})(0.674s)^2$
$-1.179m=v(0.674s)+1.179m$
$v(0.674s)=-2.36m$
$v=-3.5\frac{m}{s}$

Thanks for your help, again, haru. You've been a big help, and I honestly would have no bloody clue how I would have done this problem on my own. Thanks.