Expression for magnitude of the constant friction force

[Racoon5]

Homework Statement

After the skier reaches the ground at point P in Figure 1, she begins braking and comes to a halt at point Q. Find an expression for the magnitude of the constant friction force that acts on her between point P and Q.

Relevant Equations

W=f*d

I approached the problem using the work done by force equation (W=F*d)
In my understanding all potential energy would have been converted into kinetic energy (KE) by point P (no friciton)
We know d= L ; W= W_f ; f = F_f height = (H+h)
So the Energy at point P is entirely kinetic:
Which translates into: E_P= mg(H+h)

because W_f= mg(H+h)

so mg(H+h)= F_fx L

after solving for F_f I get F_f = mg(H+h) / L

My concern is that visually point P still has potential energy. However, the text indicates that this is not the case as the "skier reaches the ground at point P".

 

[PeroK]

Can we see the diagram? Note that the ground, expecially in a ski resort, is not necessarily flat.

 

[Racoon5]

Can we see the diagram? Note that the ground, expecially in a ski resort, is not necessarily flat.

Apologies, I forgot to attach the diagram. I have amended my original post

 

[PeroK]

Apologies, I forgot to attach the diagram. I have amended my original post

Your expression is correct. You don't have to worry about the intermediate state of kinetic + potential at point P. All that matters is the total change in PE between the start and point Q. (As the skier is at rest - has no KE - at both these points.)

 

[Racoon5]

Your expression is correct. You don't have to worry about the intermediate state of kinetic + potential at point P. All that matters is the total change in PE between the start and point Q. (As the skier is at rest - has no KE - at both these points.)

Okay if we think about it that way, is this approach correct:

##W_f = f*d##
W_f = PE = mg(H+h)
f = F_f (friction force)
d = L

so

##mg(H+h) = F_f * L##

solved for F_f =

##F_f = \frac {mg(H+h} {L}## ?

Im also struggling to use LaTeX engine as you can tell

W_f = F_f L

 

[PeroK]

Okay if we think about it that way, is this approach correct:

##W_f = f*d##
W_f = PE = mg(H+h)
f = F_f (friction force)
d = L

so

##mg(H+h) = F_f * L##

solved for F_f =

##F_f = \frac {mg(H+h} {L}## ?

Im also struggling to use LaTeX engine as you can tell

Yes, although it's not necessary always to use $d$ for the distance. In this case, you can simply write:
$$W_f = F_f L$$

 

[haruspex]

##F_f = \frac {mg(H+h} {L}## ?

Im also struggling to use LaTeX engine as you can tell

It is because you embedded [] controls in the LaTeX (for SUB, in this case).
Using the LaTeX for subscript (underscore) gives
$F_f = \frac {mg(H+h} {L}$.
The other error is a missing ). Fixing that:
$F_f = \frac {mg(H+h)} {L}$.

 

[Lnewqban]

So the Energy at point P is entirely kinetic:
Which translates into: E_P= mg(H+h)

Welcome, @Racoon5 !
That equation is incorrectly including an energy that gravity has not given away yet when the skier reaches point P, which is the height differential between points P and Q times mg.

My concern is that visually point P still has potential energy. However, the text indicates that this is not the case as the "skier reaches the ground at point P".

In this case, reaching the ground means commencement of friction action, rather than reaching the height of lowest potential energy.

Good having you here!