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So this is a conceptual question from Giancoli. It's not homework. I'm trying to understand whether I have an error in my way of thinking about this.

The problem is the following: two objects $a,b$ with $m_a=4 kg$ and $m_b=2 kg $ move with $v_a = 2 \frac{m}{s}$ and $v_b=4 \frac{m}{s}$ Now this yields that $b$ has twice the kinetic energy. The question then stated the following: What is the breaking distance of each one related to the other?

My reasoning is that $b$ has twice the kinetic Energy thus also has twice the breaking distance compared to $a$. However this is not correct.

The solution states this: "The 2-kg mass travels greater than twice as far."

This seems to contradict the Energy preservation? Why is my solution: "The 2-kg mass travels twice as far as the 4-kg mass before stopping." incorrect?

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    $\begingroup$ Take into account that the friction that accounts for braking is proportional to the normal force (in this case gravity). $\endgroup$
    – NDewolf
    Commented Dec 22, 2020 at 16:38
  • $\begingroup$ The friction is taken into account in one case but not the other, the resulting answer is the same from the book... I edited an image of the question so you understand what I mean... $\endgroup$
    – kompoloi
    Commented Dec 22, 2020 at 16:42
  • $\begingroup$ I just realized my mistake. Excuse my brainletism. YOu were right NDewolf $\endgroup$
    – kompoloi
    Commented Dec 22, 2020 at 16:49
  • $\begingroup$ @kompoloi aren't you saying the same as the hint given... ? :) $\endgroup$
    – Ankit
    Commented Dec 22, 2020 at 17:00
  • $\begingroup$ In the second one, $$\frac{1}{2}mv^2=\mathbf{F}\cdot \mathbf{s}=\mu mg s\Rightarrow s \propto v^2$$ Why this lead to wrong answer? $\endgroup$
    – Himanshu
    Commented Dec 22, 2020 at 17:02

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In the first case, when all you are given is a horizontal retarding force that is the same for both objects, the stopping distance for either object is proportional to its kinetic energy. The second mass travels twice as much as the first mass before stopping, and your reasoning would work there; no discussion of the masses themselves is actually necessary here if you know their kinetic energies.

The second case is different, however, since we are given friction: now the masses themselves are relevant, since kinetic friction depends on mass. Say $\mu_k$ is the coefficient of kinetic friction between the masses and the surface; $m_1$ and $m_2$ are their masses and $K_1$ and $K_2$ are their K.E.s.

The frictional force $-\mu_k\,m_i\,g$ does $-\mu_k\,m_i\,g\, d_i$ work on the $i^\mathrm{th}$ mass, where $d_i$ is its stopping distance. Equating this with the kinetic energy as the $i^\mathrm{th}$ mass comes to rest, we find that

$$d_i = \frac{K_i}{\mu_k\, m_i\,g}.$$ The ratio of the stopping distances is then $$\frac{d_2}{d_1} = \left(\frac{K_2}{K_1}\right) \frac{m_1}{m_2}= 2 \times 2 = 4 > 2.$$ Hence, the two-kilogram mass travels 4 times as far before stopping than the two-kilogram mass.

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  • $\begingroup$ Thank you. Very clear explanation. $\endgroup$
    – kompoloi
    Commented Dec 22, 2020 at 17:17

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