[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 $\left(g=3.7 \mathrm{~m} / \mathrm{s}^{2}\right)$ ?
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: $\frac{1}{2} k x^{2} = m g h + \mu m g d$, 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: $\frac{1}{2} k x^{2} = g h + \mu g d$.

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

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: $d_{M} = \frac{2 * \frac{1}{2} k x^{2}}{g_{M} + 2 \mu g_{M}}$.

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

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