Mechanics:puck and incline

  • #1
1. Homework Statement
A student kicks a puck with initial speed [itex] v_0 [/itex] so that it slides straight up a plane that is inclined at an angle [itex] \theta [/itex] above the horizontal. the incline has a coefficient of friction (both static and kinetic) of [itex] \mu [/itex]
Write down Newton's second law for the puck and solve it to give it's position as a function of time.

2. Homework Equations

[itex] F=m\ddot r [/itex]

The Attempt at a Solution


idk why im having such a hard time with this. imagine pucked is kicked diagonally up and to the right. that incline will be the positive x axis with angle [itex] \theta [/itex] to the horizontal.
forces
1. friction: -x direction
2. normal: positive y axis
3. gravity: straight down(component of x and y)
4. [itex] v_0 [/itex]: positive x direction

[itex] F_x=m\ddot r [/itex]
[itex] \vec F_g \sin\theta + v_0 -\mu_x = m\ddot x [/itex]

[itex] F_y=0 [/itex]
[itex] \vec N - \mu_y + \vec F_g\cos\theta=0 [/itex]

thats as far as ive gotten. are these equations correctly written?

thanks
 

Answers and Replies

  • #2
Nathanael
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A student kicks a puck with initial speed [itex] v_0 [/itex] so that it slides straight up a plane
imagine pucked is kicked diagonally up and to the right.
As I understand it, the bold part of the first quote means that the puck is not kicked diagonally up the plane.
 
  • #3
As I understand it, the bold part of the first quote means that the puck is not kicked diagonally up the plane.
after that it says it's an inclined plane.
 
  • #4
Nathanael
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after that it says it's an inclined plane.
Sorry, I thought you were thinking the puck was moving diagonally up the inclined plane instead of straight up the inclined plane.

Your equations aren't correct but it's hard to give you any guidance since you didn't explain them. Perhaps you could start by explaining what you mean by [itex]\mu_x[/itex]and [itex]\mu_y[/itex]?
[itex]\mu[/itex] is the coefficient of friction, a scalar, it doesn't have x and y components.
 
  • #5
haruspex
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##\vec F_g \sin\theta + v_0 -\mu_x = m\ddot x##
Apart from the question Nathanael asked of what ##\mu_x ## is supposed to be, the ##v_0## makes no sense there. You can't add a force to a velocity.
 
  • #6
okay...so taking out the [itex] v_0 \text { and } \mu_x , \mu_y [/itex] the equations are correct? but how do the spring constants come into play?
 
  • #7
haruspex
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okay...so taking out the [itex] v_0 \text { and } \mu_x , \mu_y [/itex] the equations are correct? but how do the spring constants come into play?
You can't just throw them away. You need to replace them with the correct terms, involving friction.
Where have spring constants been mentioned in this thread?
 

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