Find the distance traveled by a block pushed up a slope.

[Eclair_de_XII]

Homework Statement

"A block is projected up a frictionless inclined plane with initial speed v₀ = 3.50 m/s. The angle of incline is θ = 32.0°. (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

F_g = mg⋅sinθ
F_x = -F_g
Assume: arbitrary mass of block is 1-kg

The Attempt at a Solution

(b) F_g = (1kg)(9.8m/s²)(sin (-32°)) = -5.1932N
5.1932N = (1kg)(3.5m/s)/t
t = (3.5 kg⋅m/s)/5.1932N = 0.674 s

I've only figured out how to find the time it takes for the block to reach that point on the slope before it slips back down. The book says its speed when it gets back to the bottom is 3.50 m/s; same as the initial velocity. I have a vague idea how to solve for how far the block goes. Basically, I figure that once v₀ = 0, it has reached the maximum point it can reach at the initial velocity at 3.50 m/s before it comes down due to gravity, anyway. At that point, gravitational force and the force pushing the box should be equal. But I have no idea how I would apply this.

 

[cnh1995]

You don't need mass here. Just use basic kinematical equations. You have initial velocity, acceleration and final velocity. Which kinematical eauation connects these three?

 

[Eclair_de_XII]

It's not in my book, but I think it's: "d = v₀t + ½at²"? Okay, I've just done it...

d = (3.5 m/s)(0.674 s) + ½[(0 m/s - 3.5 m/s)/(0.674 s - 0 s)](0.674 s)²
= (2.359 m) + ½(-5.1932 m/s²)(0.674 s)^{2
= 2.359 m - 1.180 m = 1.18 m}

Now why is the velocity as the block slides down the same as the initial velocity, can I ask?

 

[cnh1995]

Right. I expected v²=v₀²+2ad, without involving time parameter. But since time is also asked, your method is right.

Now why is the velocity as the block slides down the same as the initial velocity, can I ask?

The block starts sliding down from rest i.e. v₀=0. You need to use the exact same equation to get final velocity now(part c).

 

[Eclair_de_XII]

To plug in acceleration, though, I would need the final velocity, which is the whole point of the equation, no?

 

[cnh1995]

To plug in acceleration, though, I would need the final velocity, which is the whole point of the equation, no?

You are asked to find the final velocity. You do have acceleration. What is pulling the block and making it slide downward?

 

[Eclair_de_XII]

What is pulling the block and making it slide downward?

Oh, gravity.
a = sin(-32°)(9.8 m/s²) = -5.1932 m/s²
v² = v₀²+2ad
= (0)² + 2(|-5.1932| m/s²)(1.18 m)
v² = (12.26 m²/s²)
√v² = √(12.26 m²/s²)
v = 3.5 m/s

Additionally, how would I go about making my equations look something like: https://upload.wikimedia.org/math/f/9/1/f9160e375f1f6fc949a99985fb1e6a72.png? I've seen it done, and I want to know if there's a guide that tells me how to do that.

 

[cnh1995]

Additionally, how would I go about making my equations look something like: https://upload.wikimedia.org/math/f/9/1/f9160e375f1f6fc949a99985fb1e6a72.png? I've seen it done, and I want to know if there's a guide that tells me how to do that.

It's calculus (vector calculus, precisely). Do you know anything about derivative of a function?

 

[Eclair_de_XII]

No, I meant the font styling. How do I change it?

 

[cnh1995]

No, I meant the font styling. How do I change it?

Oh! I thought you wanted to derive kinematical equations using calculus.. I am not sure but it's LaTex. You'll find the LaTex/BBcode guide under the editor space here (below "have something to add?" space).

 

[Eclair_de_XII]

Found it. Thanks!

https://www.physicsforums.com/help/latexhelp/