Answer :
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.
