Need help figuring out this problem

[Karhitect]

Homework Statement

6. A stunned Batman finds himself in a vault on a ledge. In an attempt the trap him the Joker runs out of the vault while the door closes. Batman fires a grappling hook into the ceiling at an angle, θ, attached to a line of length, L , and manages to pull himself off the ledge swinging down to the floor below. At the very bottom of the swing he let's go of the line to the grappling hook in an attempt to slide the rest of the way out of the door. Find an equation for how far Batman slides, x, as a function of L, θ and μk, the coefficient of kinetic friction between Batman and the floor. If the door is 9.00 meters away and the length of the line is 6.00 meters, the angle 60 degrees and the coefficient of friction 0.250 did Batman make it out? If he did, how fast was he sliding when he went through the door?

Homework Equations

f_k (x)=E₁ -E₂
started out with that equation

The Attempt at a Solution

E1=mgL(1-cosθ)
E2=1/2mv²
fk=μkmg
so,

μkmg(x)=mgL(1-cosθ)-1/2mv²

then,
x=L(1-cosθ)/μk - v² /2μkg


thats what i got for the equation but I am not sure its right because when i try to use it to find how far he slides i can't because i don't know what v is. I am really lost.

 

[TSny]

"How far Batman slides" means "How far Batman slides before coming to rest."

 

[Karhitect]

yea still lost lol

 

[TSny]

Batman slides until he comes to rest. What is the value of v when an object is at rest? So, what is the value of v that you should use in your equation?

 

[tiny-tim]

Welcome to PF!

Hi Karhitect! Welcome to PF!

E1=mgL(1-cosθ)

fine so far … that's the kinetic energy at the bottom of the swing

E2=1/2mv²
fk=μkmg
so,

μkmg(x)=mgL(1-cosθ)-1/2mv²

then,
x=L(1-cosθ)/μk - v² /2μkg

this is rather confusing …

what is your E₂ supposed to be? ​

if it's the energy at the start of the slide, then it's just E₁, and v isn't relevant

if it's the energy at the door, then v is the answer

 

[TSny]

You should find that Batman slides more than 9.0 m before coming to rest. Therefore, he will make it out. If you want to know his speed as he goes through the door, what value would you use for x and what symbol would you solve for using your equation?

 

[Karhitect]

ohh duh lol that makes sense...i found that he slides 12 m ?

i think you would use 9m for x

would i use my same equation only solving for v this time?

 

[TSny]

Yes, 12 m. And you have the right idea for finding v at the door.

 

[Karhitect]

ok so then i got vf= mgL(1-cosθ)-μkmg

and i got 7.35 m/s ??

 

[TSny]

You have x = L(1-cosθ)/μ_k - v² /2μ_kg

Rearrange: v² /2μ_kg = L(1-cosθ)/μ_k - x

You'll still need to multiply through by 2μ_kg and then take a square root.

 

[Karhitect]

i think i did something wrong because i got a negative number...when i brought the 2μkg over i brought it next to the x...then i square rooted the whole thing. but because i got a negative number i couldn't square root...

 

[TSny]

v² /2μ_kg = L(1-cosθ)/μ_k - x

When you multiply both sides by 2μ_kg, the whole right hand side gets multiplied.

v² = 2μ_kg [L(1-cosθ)/μ_k - x]

 

[Karhitect]

ohhh i see i was thinking that you had to minus the whole thing by x...soooo i got 3.83m/s

 

[TSny]

Looks Good!

 

[Karhitect]

Thank you sooo much!