# Does this problem contradict conservation of energy?

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? • Take into account that the friction that accounts for braking is proportional to the normal force (in this case gravity). Dec 22, 2020 at 16:38
• 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... Dec 22, 2020 at 16:42
• I just realized my mistake. Excuse my brainletism. YOu were right NDewolf Dec 22, 2020 at 16:49
• @kompoloi aren't you saying the same as the hint given... ? :) Dec 22, 2020 at 17:00
• 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? Dec 22, 2020 at 17:02

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.