- #1
ahero4eternity
- 10
- 0
[SOLVED] Work-Energy Principle
1. The problem...
On an essentially frictionless horizontal ice-skating rink, a skater moving at 3.00 m/s encounters a rough patch that reduces her speed by 45.0 % to a friction force that is 25.0 % of her weight.
Use the work-energy principle to find the length of the rough patch.2. The relevant equation...
KEi + PEi + W(by friction) = KEf + PEf where i = initial and f = final3. My attempt...
Since Work by friction = -Fd, then you can substitute F for (.25 x mg), correct? that is because the Force of friction (F) is 25% of her weight (mg). Ultimately, this is going to allow m to cancel out. With that, I have the equation:
.5mv^2 + mgh + (-.25mg)d = .5mv^2 + mgh
Factoring out the m so it can cancel out I now have:
m (.5v^2 + gh + (-.25g)d) = m (.5v^2 + gh)
Since this is all happening on a level surface, then h = 0. Now I am left with:
.5v^2 + (-.25g)d = .5vf^2
Now, plug in our known values...
.5(3^2) + (-.25 x 9.8)d = .5(.45 x 3)^2
Note: v final = 3 x 45% as stated in the problem
Solving for d, I get 1.46 meters. But this is wrong...
I've also tried this same exact setup using a final speed of 0, getting a distance of 1.84 meters. That, too, is wrong.
1. The problem...
On an essentially frictionless horizontal ice-skating rink, a skater moving at 3.00 m/s encounters a rough patch that reduces her speed by 45.0 % to a friction force that is 25.0 % of her weight.
Use the work-energy principle to find the length of the rough patch.2. The relevant equation...
KEi + PEi + W(by friction) = KEf + PEf where i = initial and f = final3. My attempt...
Since Work by friction = -Fd, then you can substitute F for (.25 x mg), correct? that is because the Force of friction (F) is 25% of her weight (mg). Ultimately, this is going to allow m to cancel out. With that, I have the equation:
.5mv^2 + mgh + (-.25mg)d = .5mv^2 + mgh
Factoring out the m so it can cancel out I now have:
m (.5v^2 + gh + (-.25g)d) = m (.5v^2 + gh)
Since this is all happening on a level surface, then h = 0. Now I am left with:
.5v^2 + (-.25g)d = .5vf^2
Now, plug in our known values...
.5(3^2) + (-.25 x 9.8)d = .5(.45 x 3)^2
Note: v final = 3 x 45% as stated in the problem
Solving for d, I get 1.46 meters. But this is wrong...
I've also tried this same exact setup using a final speed of 0, getting a distance of 1.84 meters. That, too, is wrong.