Explore Westonci.ca, the leading Q&A site where experts provide accurate and helpful answers to all your questions. Discover solutions to your questions from experienced professionals across multiple fields on our comprehensive Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
I'll bite:
-- Since the sled's mass is 'm', its weight is 'mg'.
-- Since the coefficient of kinetic friction is μk, the force acting opposite to the direction it's sliding is (μk) times (mg) .
-- If the pulling force is constant 'F', then the horizontal forces on the sled
are 'F' forward and (μk · mg) backwards.
-- The net force on the sled is (F - μk·mg).
(I regret the visual appearance that's beginning to emerge,
but let's forge onward.)
-- The sled's horizontal acceleration is (net force) / (mass) = (F - μk·mg) / m.
This could be simplified, but let's not just yet.
-- Starting from rest, the sled moves a distance 's' during time 't'.
We know that s = 1/2 a t² , and we know what 'a' is. So we can write
s = (1/2 t²) (F - μk·mg) / m .
Now we have the distance, and the constant force.
The total work is (Force x distance), and the power is (Work / time).
Let's put it together and see how ugly it becomes. Maybe THEN
it can be simplified.
Work = (Force x distance) = F x (1/2 t²) (F - μk·mg) / m
Power = (Work / time) = F (t/2) (F - μk·mg) / m
Unless I can come up with something a lot simpler, that's the answer.
To simplify and beautify, make the partial fractions out of the
2nd parentheses:
F (t/2) (F/m - μk·m)
I think that's about as far as you can go. I tried some other presentations,
and didn't find anything that's much simpler.
Five points,ehhh ?
The power developed by the force in the sled is [tex]\boxed{\frac{{Ft}}{2}\left({\frac{F}{m}-{\mu _k}g}\right)}[/tex] .
Further Explanation:
The Sled is being pulled on a rough surface with the coefficient of kinetic friction [tex]{\mu _k}[/tex] with a force of magnitude [tex]F[/tex] .
The expression for the force balancing on the sled is written as:
[tex]F-{\mu _k}N=ma[/tex]
Here, [tex]N[/tex] is the normal reaction force on the sled, [tex]m[/tex] is the mass of the sled and [tex]a[/tex] is the acceleration of the sled due to the force.
The normal reaction of the surface acting on the sled is:
[tex]N=mg[/tex]
Therefore, the acceleration of the sled is:
[tex]\begin{gathered}F-{\mu _k}mg=ma\hfill\\a=\frac{{F-{\mu _k}mg}}{m}\hfill\\\end{gathered}[/tex]
Since the sled starts from s=rest and the force acts on the sled for time [tex]t[/tex] . So, the distance covered by the sled in time [tex]t[/tex] is:
[tex]S=ut+\frac{1}{2}a{t^2}[/tex]
Here, [tex]S[/tex] is the distance covered by the sled in time [tex]t[/tex] and [tex]u[/tex] is the initial velocity of the sled.
Substitute [tex]0[/tex] for [tex]u[/tex] and [tex]\frac{{F-{\mu _k}mg}}{m}[/tex] for [tex]a[/tex] in above expression.
[tex]S=\frac{1}{2}\left({\frac{{F-{\mu _k}mg}}{m}}\right){t^2}[/tex]
The work done by the force in making the sled move on the surface is given as:
[tex]W = F \times S[/tex]
Substitute the value of [tex]S[/tex] in above expression.
[tex]\begin{aligned}W&=F\times\frac{1}{2}\left({\frac{{F-{\mu _k}mg}}{m}}\right){t^2}\\&=\frac{{F{t^2}}}{2}\left({\frac{{F-{\mu _k}mg}}{m}}\right)\\\end{aligned}[/tex]
The power developed by the force acting on the sled is:
[tex]P = \frac{W}{t}[/tex]
Substitute [tex]\frac{{F{t^2}}}{2}\left({\frac{{F-{\mu _k}mg}}{m}}\right)[/tex] for [tex]W[/tex] in above expression.
[tex]\begin{aligned}P&=\frac{{\frac{{F{t^2}}}{2}\left({\frac{{F-{\mu _k}mg}}{m}}\right)}}{t}\\&=\frac{{Ft}}{2}\left({\frac{F}{m}-{\mu _k}g}\right)\\\end{aligned}[/tex]
Thus, the power developed by the force in the sled is [tex]\boxed{\frac{{Ft}}{2}\left({\frac{F}{m}-{\mu _k}g}\right)}[/tex] .
Learn More:
1. Choose the 200 kg refrigerator. Set the applied force to 400 n https://brainly.com/question/4033012
2. It's been a great day of new, frictionless snow. Julie starts at the top of the 60∘ https://brainly.com/question/3943029
3. If the coefficient of static friction between a table and a uniform massive rope is μs, https://brainly.com/question/2959748
Answer Details:
Grade: High School
Subject: Physics
Chapter: Friction
Keywords:
Sled, pulled horizontally, coefficient of kinetic friction, mu_k, moves a distance, constant horizontal force, average power, Pavg, created by force.
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.
