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Assume that friction causes the car to lose 37,800 joules of energy as it travels from the top to the bottom of the first hill. What is the car's velocity at the bottom of the first hill?

Use the following formula to calculate the velocity:

[tex]\[
v = \sqrt{\frac{2(E_{\text{initial}} - E_{\text{friction}})}{m}}
\][/tex]

Sagot :

GinnyAnswer
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].
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