Sure, let's tackle this problem step-by-step.
### Step 1: Calculate the Potential Energy at the Top of the Hill
We start by calculating the potential energy at the top of the hill.
The formula for potential energy [tex]\(PE\)[/tex] is:
[tex]\[
PE = m \cdot g \cdot h
\][/tex]
where:
- [tex]\(m\)[/tex] is the mass of the car,
- [tex]\(g\)[/tex] is the acceleration due to gravity ([tex]\(9.81 \, \text{m/s}^2\)[/tex]),
- [tex]\(h\)[/tex] is the height of the hill.
Assuming the mass [tex]\(m\)[/tex] of the car is [tex]\(1 \, \text{kg}\)[/tex] and the height [tex]\(h\)[/tex] of the hill is [tex]\(1 \, \text{m}\)[/tex]:
[tex]\[
PE = 1 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 1 \, \text{m} = 9.81 \, \text{J}
\][/tex]
### Step 2: Determine the Total Energy Lost Due to Friction
We are given that the car loses [tex]\(37,800\)[/tex] joules of energy due to friction.
### Step 3: Calculate the Kinetic Energy at the Bottom of the Hill
The kinetic energy ([tex]\(KE\)[/tex]) at the bottom of the hill can be calculated by subtracting the energy lost due to friction from the initial potential energy.
[tex]\[
KE = PE - \text{Energy lost due to friction}
\][/tex]
Substituting the values:
[tex]\[
KE = 9.81 \, \text{J} - 37,800 \, \text{J} = -37,790.19 \, \text{J}
\][/tex]
Since it is not possible to have a negative kinetic energy, the kinetic energy at the bottom of the hill is [tex]\(0 \, \text{J}\)[/tex].
### Step 4: Calculate the Velocity at the Bottom of the Hill
Kinetic energy is also given by the formula:
[tex]\[
KE = \frac{1}{2} m v^2
\][/tex]
Solving for velocity [tex]\(v\)[/tex]:
[tex]\[
v = \sqrt{\frac{2 KE}{m}}
\][/tex]
Since the kinetic energy at the bottom is [tex]\(0 \, \text{J}\)[/tex]:
[tex]\[
v = \sqrt{\frac{2 \times 0 \, \text{J}}{1 \, \text{kg}}} = \sqrt{0} = 0 \, \text{m/s}
\][/tex]
### Conclusion
The potential energy at the top of the hill is [tex]\(9.81 \, \text{J}\)[/tex], the kinetic energy at the bottom of the hill is [tex]\(-37,790.19 \, \text{J}\)[/tex], and the car's velocity at the bottom is [tex]\(0 \, \text{m/s}\)[/tex].
Thus, the car's velocity at the bottom of the first hill is [tex]\(0 \, \text{m/s}\)[/tex].
