Solving v^2/2=gs(sin(α)+cos(α)*k): Get k with Steps

In summary: k*cos(α)*g*s = (1/2)*v2-g*sin(α)*sk = v2-g*sin(α)*s / 2*cos(α)*g*sv2/2 = g*sin(α)*s + k*cos(α)*g*sk*cos(α)*g*s = 1/2*v2-g*sin(α)*sk = v2-g*sin(α)*s / 2*cos(α)*g*s
  • #1
CarrotWilliams
4
1
I've gotten to this point:
v^2/2=gs(sin(α)+cos(α)*k)

I'm suppost to get "k" from this equation can some help by showing me the steps because I'm a bit confused on how to do it.
 
Physics news on Phys.org
  • #2
CarrotWilliams said:
I've gotten to this point:
v^2/2=gs(sin(α)+cos(α)*k)

I'm suppost to get "k" from this equation can some help by showing me the steps because I'm a bit confused on how to do it.
Welcome to the PF.

In your future schoolwork posts here, please fill out the Homework Help Template that you are provided, and show your best efforts to work the problem.

So start off by distributing terms on the righthand side (RHS) -- you should separate the two terms. Then move the first term on the RHS to the LHS with subtraction. Then how do you get k all by itself on the RHS?
 
  • #3
berkeman said:
Welcome to the PF.

In your future schoolwork posts here, please fill out the Homework Help Template that you are provided, and show your best efforts to work the problem.

So start off by distributing terms on the righthand side (RHS) -- you should separate the two terms. Then move the first term on the RHS to the LHS with subtraction. Then how do you get k all by itself on the RHS?
So I've gotten the result: k= v^2-g*sin(α)*s/2*cos(α)*g*s <----- is this correct ?
 
  • #4
CarrotWilliams said:
So I've gotten the result: k= v^2-g*sin(α)*s/2*cos(α)*g*s <----- is this correct ?
Could you show each of your steps? That would make it easier to check... :smile:
 
  • #5
Also, please be careful and explicit with parenthesis when typing the equations out with just text. There is a LaTeX primer under INFO, Help-How-To at the top of the page, BTW. :smile:
 
  • #6
berkeman said:
Could you show each of your steps? That would make it easier to check... :smile:
  1. v2/2 = g*sin(α)*s + k*cos(α)*g*s
  2. k*cos(α)*g*s = 1/2*v2-g*sin(α)*s
  3. k = v2-g*sin(α)*s / 2*cos(α)*g*s
 
  • #7
CarrotWilliams said:
  1. v2/2 = g*sin(α)*s + k*cos(α)*g*s
  2. k*cos(α)*g*s = 1/2*v2-g*sin(α)*s
  3. k = v2-g*sin(α)*s / 2*cos(α)*g*s
Looks okay to me, but as I mentioned, it's best to explicity show parenthesis when just typing out equations in text. So I'd modify your last equation to:

k = [ v2-g*sin(α)*s ] / 2*cos(α)*g*s
 
  • #8
BTW, there are also other ways to express the final RHS, depending on what you want to do with it going forward. For example, you could show it as two separate fractions, with a tan(α) in the 2nd term... :smile:
 
  • #9
berkeman said:
Looks okay to me, but as I mentioned, it's best to explicity show parenthesis when just typing out equations in text. So I'd modify your last equation to:

k = [ v2-g*sin(α)*s ] / 2*cos(α)*g*s

ok thanks for your help !
 
  • Like
Likes berkeman
  • #10
CarrotWilliams said:
  1. v2/2 = g*sin(α)*s + k*cos(α)*g*s
  2. k*cos(α)*g*s = 1/2*v2-g*sin(α)*s
  3. k = v2-g*sin(α)*s / 2*cos(α)*g*s
You do need a set of parentheses around the denominator, to remove ambiguity. White space around the division symbol does not change its meaning in the real world, even though some middle high teachers may have misled you otherwise.

So your eqn 3 needs to be written as k = (v2-g*sin(α)*s) / (2*cos(α)*g*s)

For exactly the same reason, I'd like to see your eqn 2 written with parentheses around the 1/2 to emphasise clarity, e.g.,
k*cos(α)*g*s = (1/2)*v2-g*sin(α)*s

But it's still not right. You messed up in going from 2. to 3, probably because of this sloppiness with the solidus sign, where you have it doing one thing in eqn 2 and something different in 3.

Try that step again, starting with your eqn 2.
 
  • Like
Likes berkeman

Related to Solving v^2/2=gs(sin(α)+cos(α)*k): Get k with Steps

1. How do I solve the equation v^2/2=gs(sin(α)+cos(α)*k) to get k?

To solve this equation and find the value of k, you will need to follow these steps:

  1. Isolate the variable k by subtracting gs(sin(α)) from both sides.
  2. Divide both sides by cos(α) to isolate k.
  3. This will give you the value of k and allow you to solve the equation.

2. What do the variables v, g, α, and k represent in this equation?

The variable v represents the initial velocity, g represents the acceleration due to gravity, α represents the angle of the incline or ramp, and k represents a constant value.

3. Can you provide an example problem using this equation?

Sure, let's say we have an object sliding down a ramp with an initial velocity of 10 m/s at an angle of 30 degrees. The acceleration due to gravity is 9.8 m/s^2. Plugging these values into the equation, we get 100/2 = 9.8(sin(30) + cos(30)*k). Solving for k, we get a value of approximately 0.58.

4. How is this equation used in physics?

This equation is used in physics to calculate the coefficient of friction between an object and a surface. The value of k represents the coefficient of friction and can be used to determine how much resistance the surface provides to the sliding object.

5. Is there a simpler way to solve this equation?

Yes, depending on the specific problem, there may be other equations or methods that can be used to solve for k. It is always important to consider the given variables and the goal of the problem before choosing an appropriate method for solving.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
881
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
281
  • Introductory Physics Homework Help
Replies
33
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
0
Views
607
  • Introductory Physics Homework Help
Replies
3
Views
800
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
Back
Top