Question

A sled of mass m is being pulled horizontally by a constant horizontal force of magnitude F. The coefficient of kinetic friction is μk. During time interval t, the sled moves a distance s, starting from rest.
Find the average power Pavg created by the force F.

Answer 1

The power developed by the force in the sled is $\boxed{\frac{{Ft}}{2}\left({\frac{F}{m}-{\mu _k}g}\right)}$ .

Further Explanation:

The Sled is being pulled on a rough surface with the coefficient of kinetic friction ${\mu _k}$  with a force of magnitude $F$ .

The expression for the force balancing on the sled is written as:

$F-{\mu _k}N=ma$

Here, $N$  is the normal reaction force on the sled, $m$  is the mass of the sled and $a$  is the acceleration of the sled due to the force.

The normal reaction of the surface acting on the sled is:

$N=mg$

Therefore, the acceleration of the sled is:

$\begin{gathered}F-{\mu _k}mg=ma\hfill\\a=\frac{{F-{\mu _k}mg}}{m}\hfill\\\end{gathered}$

Since the sled starts from s=rest and the force acts on the sled for time $t$ . So, the distance covered by the sled in time $t$  is:

$S=ut+\frac{1}{2}a{t^2}$

Here, $S$  is the distance covered by the sled in time $t$  and   $u$ is the initial velocity of the sled.

Substitute $0$  for $u$   and $\frac{{F-{\mu _k}mg}}{m}$  for $a$  in above expression.

$S=\frac{1}{2}\left({\frac{{F-{\mu _k}mg}}{m}}\right){t^2}$

The work done by the force in making the sled move on the surface is given as:

$W = F \times S$

Substitute the value of $S$  in above expression.

$\begin{aligned}W&=F\times\frac{1}{2}\left({\frac{{F-{\mu _k}mg}}{m}}\right){t^2}\\&=\frac{{F{t^2}}}{2}\left({\frac{{F-{\mu _k}mg}}{m}}\right)\\\end{aligned}$

The power developed by the force acting on the sled is:

$P = \frac{W}{t}$

Substitute $\frac{{F{t^2}}}{2}\left({\frac{{F-{\mu _k}mg}}{m}}\right)$   for $W$  in above expression.

$\begin{aligned}P&=\frac{{\frac{{F{t^2}}}{2}\left({\frac{{F-{\mu _k}mg}}{m}}\right)}}{t}\\&=\frac{{Ft}}{2}\left({\frac{F}{m}-{\mu _k}g}\right)\\\end{aligned}$

Thus, the power developed by the force in the sled is $\boxed{\frac{{Ft}}{2}\left({\frac{F}{m}-{\mu _k}g}\right)}$ .

Learn More:

1. Choose the 200 kg refrigerator. Set the applied force to 400 n brainly.com/question/4033012

2. It's been a great day of new, frictionless snow. Julie starts at the top of the 60∘ brainly.com/question/3943029

3. If the coefficient of static friction between a table and a uniform massive rope is μs, 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.

Answer 2

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 ?