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

1. Mar 10, 2016

### Eclair_de_XII

1. The problem statement, all variables and given/known data
"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?"

2. Relevant equations

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

3. 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.

2. Mar 10, 2016

### 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.

3. Mar 10, 2016

### 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.

4. Mar 10, 2016

### haruspex

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

5. Mar 10, 2016

### 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 $-9.8\frac{m}{s^2}$.

6. Mar 10, 2016

### 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?

7. Mar 10, 2016

### 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°.

8. Mar 10, 2016

### haruspex

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

9. Mar 10, 2016

### 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.

10. Mar 10, 2016

### haruspex

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?

11. Mar 10, 2016

### 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.

12. Mar 10, 2016

### cnh1995

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.

13. Mar 10, 2016

### haruspex

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

14. Mar 10, 2016

### haruspex

The angle between what and what is 180?

15. Mar 10, 2016

### Eclair_de_XII

The front and the back?

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

16. Mar 10, 2016

### cnh1995

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

17. Mar 10, 2016

### haruspex

18. Mar 10, 2016

### 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.

19. Mar 10, 2016

### haruspex

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

20. Mar 10, 2016

### 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?