Energy/Work Problem: Friction of sliding down pole

[BBA Biochemistry]

Homework Statement

A fireman of mass m slides a distance d down a pole. He starts from rest. He moves as fast at the bottom as if he had stepped off a platform a distance h≤d above the ground and descended with negligible air resistance.What average friction force did the fireman exert on the pole?
Express your answer in terms of the variables m, d, h, and appropriate constants.

Homework Equations

U=mgh
K= 0.5mv^2
K1+U1+W(other)=K2+U2
W=Fs

The Attempt at a Solution

Since friction is a nonconservative force, I figure I'm supposed to use K1+U1+W(other)=K2+U2 and find the W(other), and then use the equation W=Fs to find the actual force of the friction.

When the fireman is at rest at height, d, he has no KE, just PE, which is:

U1= mgd

As he reaches the ground, some of the potential energy was converted to kinetic energy, but the rest of it was lost through friction:

mgd+W(other)=K2=0.5mv^2

To find W(other) I need to find the velocity of the fireman. This is where I start to get unsure. The problem states that his velocity is constant at all heights (h) after he starts sliding down from his initial position, d. I decided to set up another conservation equation, in which he starts at some height, h, and then reaches the ground:

mgh=0.5mv^2 - W(other2)

Since W=Fs, and s=h, I substituted:

mgh=0.5mv^2 - Fh

Where F is the friction force I want to find out. I solved for v^2, getting:

v^2= (2/m) * (mgh + Fh)

I plugged this back into the conservation equation for the entire distance, getting:

W(other)=mgd-0.5m[(2/m) * (mgh + Fh)]=mgd-mgh+Fh

W(other) is also equal to a force over a distance, so:

mgd-mgh+Fh=W=Fd

Isolating F:

mg(d-h)=Fd-Fh=F(d-h)

Then the (d-h) terms cancel, getting mg=F. However, this is incorrect, and the problem states that the answer has the term d in it.

Can anyone point out where my logic and/or math was wrong?

 

[Doc Al]

As he reaches the ground, some of the potential energy was converted to kinetic energy, but the rest of it was lost through friction:

mgd+W(other)=K2=0.5mv^2

You went off the rails (a bit) when you introduced the speed "v". Instead, just express U2+ K2 in terms of the given variables.

Also, the work done by friction is negative, so be careful with signs.

 

[BBA Biochemistry]

I'm not really sure how to express U2+K2 in terms of the given variables d, h, and m.

I'm at mgd = U1 and K1=0.

So mgd + 0 + W(other) = U2 + K2.

Can I equate W(other)=F*s, making the equation:

mgd + F*d = U2 + K2 ?

If U2 equals zero (since the fireman is on the ground), that leaves:

mgd + F*d = K2.

But I still don't know K2 or F. Any hints on where I'm supposed to get my second equation for a system of equations? I haven't used the "He moves as fast at the bottom as if he had stepped off a platform a distance h≤d above the ground and descended with negligible air resistance." part of the problem yet.

 

[haruspex]

mgd + F*d = K2.

Good. Now find the corresponding equation for the case where there is no friction and the starting height is h.

 

[BBA Biochemistry]

Now find the corresponding equation for the case where there is no friction and the starting height is h.

Would that be as simple as:

mgh = K2

Since there is no initial kinetic energy nor final potential energy?

Since speed is the same at the bottom regardless of air resistance (according to the problem), would I just set these two equations equal to each other and solve for Fk?

 

[BBA Biochemistry]

I just submitted the answer and this was correct. Thanks for the help everyone!