Will Dropping His Coat & Boots Help Him Slide Faster?

[Moose100]

Homework Statement

A boy is sliding down a long icy hill on his sled. In order
to decrease his mass and increase his velocity, he drops
his heavy winter coat and heavy boots from (he sled while
he is moving. Will his strategy work?

Homework Equations

KE=1/2mv2
p=mv
mgh

The Attempt at a Solution

I thought that since he is decreasing his mass his velocity will go up. B/c in F=ma m and a are inversely related. I.e. if I drop mass a goes up which increases v?? Instead the answer (multiple choice) says that it won't work b/c his PE will go down.

How does PE and KE allow for his velocity to go down? How does this relate to momentum. I get that a reduction in energy will reduce but it's not clear how it specifically relates to velocity. Thanks!

 

[Brown Arrow]

we have KE= 1/2 M* v^2 and P=M*v

KE= (1/2)(P^2)/M

 

[Moose100]

we have KE= 1/2 M* v^2 and P=M*v

KE= (1/2)(P^2)/M

Ok but the question is specifically concerned about speed.

 

[tal444]

His potential energy will decrease as well, as PE = mgh. If he decreases his mass, his PE is decreased. Assuming no friction, PE = KE. If PE decreases, then it stands to reason that his KE will decrease, and thus his speed cannot possibly increase.

 

[Moose100]

Ok that's what I thought. I guess what I was asking is what would that look like mathematically? Or is it more conceptual?
maybe I am being too picky

 

[gash789]

Mathematically I would argue it something like:

The total energy of the system must be conserved, therefore

$\textrm{K.E} + \textrm{P.E}= \textrm{const.}$

$\Rightarrow \frac{1}{2}mv(x)^{2} + mgh(x)= \textrm{const.}$

The constant is arbitrary since we can define the potential as we like. If we define it such that the constant is zero at the point he looses weight then rearranging:

$\frac{1}{2}mv(x)^{2} + mgh(x)= 0$
$\Rightarrow v(x)^{2} = \frac{2mgh(x)}{m} =2gh(x)$

Hence the boys velocity is not dependent on his mass.

 

[Moose100]

That's what I thought too. So why is it important to consider his mass?

 

[jhae2.718]

If the total mechanical energy is independent of mass, and specific kinetic energy is v²/2, will changing his mass affect his speed? (This question is equivalent to the problem...)

 

[Moose100]

The answer to the problem says that he loses PE to the objects he leaves behind.

So he loses PE but the speed doesn't change?

 

[Moose100]

Ok, have to admit I didn't fully read through this one.

Let's examine two points in time, with the coat etc. and after they are discarded. When we change this, we go from having a total mass M to some mass m, where m < M.

Mechanical energy (KE + PE) is still conserved, but mass is changing, as is velocity and height...so if you solve for velocity, what do you get?

The kinematic equation that looks like KE but indep of mass

 

[Moose100]

v2=√(m1v1^2 +2m1gh1-2m2gh2)

 

[jhae2.718]

Ok, I think I'm leading you down the wrong direction here. I used an expression derived assuming constant mass with quantities where mass isn't constant; I'm not quite sure why I went that way other than that it's a long day and I'm tired. I deleted the previous two posts I made under that assumption so as not to mislead future readers.
---

What happens is that when the boy throws away the coat and boots, the energy of those items leaves the system. The mechanical energy within the system is still conserved, but the total amount is reduced by the energy of the coat and boots which leaves (and goes to the surroundings).

So, the velocity is still independent of the mass and thus does not increase.