Calculating Velocity Using Work and Friction

In summary, the problem involves a skier moving at 5.00 m/s encountering a long, rough horizontal patch of snow with a coefficient of kinetic friction of 0.220. The patch is 2.9 meters long and the question asks for the skier's velocity after passing the patch. The solution involves using work and avoiding Newtons, finding the final velocity using the formula v_{final} = \sqrt{ -2(-\mu_{k}gs+\frac{1}{2}v^{2}_{initial})} and correcting errors in the initial attempt at a solution.
  • #1
rsala
40
0

Homework Statement


A skier moving at 5.00 m/s encounters a long, rough horizontal patch of snow having coefficient of kinetic friction 0.220 with her skis. the patch is 2.9 meters long.
what is her velocity after passing the patch?

problem must be solved by using work, avoid Newtons

Homework Equations


work = fs
work = [tex]K_{final} - K_{initial} =- \frac{1}{2}m v^{2}_{final}- \frac{1}{2}m v^{2}_{initial}[/tex]

The Attempt at a Solution


let s be displacement (2.9m)

[tex] -\mu_{k}mg*s = - \frac{1}{2}m v^{2}_{final}- \frac{1}{2} m v^{2}_{initial}
[/tex]

m is irrelevant, factor it out and cancel.

[tex]
-\mu_{k}g*s = - \frac{1}{2} v^{2}_{final}- \frac{1}{2} v^{2}_{initial}[/tex]

solve for [tex] v_{final} [/tex]

[tex]\frac{-\mu_{k}gs+.5v^{2}_{initial}}{-.5} = v^{2}_{final}
[/tex]

[tex]
\sqrt{-2(-\mu_{k}gs+\frac{1}{2}v^{2}_{initial})} = v_{final}
[/tex]

[tex]
\sqrt{ -12.4952}
[/tex]
cannot take sqrt of negative number.
i can't go beyond this part, is there a solution?
 
Last edited:
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  • #2
[tex]
\mu_{k}g*s = \frac{1}{2} v^{2}_{initial}- \frac{1}{2} v^{2}_{final}[/tex]

Where are you finding all the extra minus signs?
 
  • #3
in which part
 
  • #4
rsala said:
in which part

All over. Look at your first expression for K_(final)-K(initial). That's supposed to be the difference of two kinetic energies, not the negative of their sum.
 
  • #5
o I've messed my formula up =/
i got correct answer now, thanks
 

Related to Calculating Velocity Using Work and Friction

1. What is the definition of work in physics?

In physics, work is defined as the amount of force applied over a distance. It is calculated by multiplying the magnitude of the force by the displacement of the object in the direction of the force.

2. How is work related to velocity?

Work and velocity are related through the concept of kinetic energy. Kinetic energy is the energy an object possesses due to its motion, and it is directly proportional to the square of its velocity. Therefore, as an object's velocity increases, its kinetic energy and the amount of work it can do also increase.

3. What is the formula for calculating work?

The formula for calculating work is W = F * d * cos(theta), where W is work, F is the magnitude of the force applied, d is the displacement of the object, and theta is the angle between the force vector and the displacement vector.

4. How can work and velocity be used to determine power?

Power is the rate at which work is done, or the amount of work done per unit of time. It can be calculated by dividing the work done by the time it takes to do that work. Therefore, by knowing the work done and the velocity at which it was done, we can determine the power of the object or system.

5. What is the difference between work and velocity problems in one dimension and two dimensions?

In one-dimensional problems, the force and displacement vectors are in the same direction, making the calculation of work simpler. In two-dimensional problems, the force and displacement vectors may be at an angle to each other, requiring the use of trigonometry to calculate the work. Additionally, two-dimensional problems may also involve considering the work done by multiple forces acting on an object.

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