[Bonus] On Earth, a mass is placed against a spring that has been
1 point compressed $48\mathrm{cm}$ as shown. The horizontal surface is frictionless, but the inclined surface is not and has a coefficient of kinetic friction of 0.4 . The spring constant is $550\mathrm{N}/\mathrm{m}$. After being released from the spring, the mass travels $2\mathrm{m}$ up the incline. How far up the incline would the mass travel if this experiment was done on Mars $(g=3.7\mathrm{m}/{\mathrm{s}}^2)$ ?
Your answer
(1)

Step 1 ：Let's denote the distance the mass would travel on Mars as $d$. The total mechanical energy of the system is conserved, so the work done by the spring equals the work done against gravity and friction. On Earth, this gives us: $\frac{1}{2}k{x}^2=mgh+\mu mgd$, where $k$ is the spring constant, $x$ is the compression of the spring, $m$ is the mass, $g$ is the acceleration due to gravity, $h$ is the height the mass is lifted, and $d$ is the distance the mass travels.

Step 2 ：Since we don't know the mass $m$, we can rearrange the equation to get rid of it: $\frac{1}{2}k{x}^2=gh+\mu gd$.

Step 3 ：Substituting the given values, we get: $\frac{1}{2}\ast 550\ast (0.48{)}^2=9.8\ast 2+0.4\ast 9.8\ast 2$.

Step 4 ：Solving this equation, we find that the left side equals $63.36$ and the right side equals $27.44$.

Step 5 ：Since the total mechanical energy is conserved, the work done by the spring on Mars should be the same as on Earth, so we have: $\frac{1}{2}k{x}^2=m{g}_M{h}_M+\mu m{g}_M{d}_M$, where $g_M$ is the acceleration due to gravity on Mars, $h_M$ is the height the mass is lifted on Mars, and $d_M$ is the distance the mass travels on Mars.

Step 6 ：Again, we can rearrange the equation to get rid of the mass: $\frac{1}{2}k{x}^2=g_M{h}_M+\mu g_M{d}_M$.

Step 7 ：Since the height the mass is lifted and the distance it travels are proportional, we can write $h_M=\frac{d_M}{2}$.

Step 8 ：Substituting this into the equation, we get: $\frac{1}{2}k{x}^2=g_M\frac{d_M}{2}+\mu g_M{d}_M$.

Step 9 ：Solving for $d_M$, we get: $d_M=\frac{2\ast \frac{1}{2}k{x}^2}{g_M+2\mu g_M}$.

Step 10 ：Substituting the given values, we get: $d_M=\frac{2\ast 63.36}{3.7+2\ast 0.4\ast 3.7}$.

Step 11 ：Solving this equation, we find that $d_M=\boxed{{\displaystyle 24.8}}$ meters.