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**Problem 3: Roller Coaster Physics**

A 300 kg roller coaster car starts with a speed of 3 m/s at a height of 18 m above the ground. The car then travels through an up-and-down path. When the car is 5 m above the ground, its speed increases to 12 m/s. The acceleration due to gravity is given as \( g = 9.8 \, \text{m/s}^2 \).

c. From the energy equation, obtain the work \( W_f \) spent on friction.

Expert Solution

Step 1

Conservation of energy

$\begin{array}{rcl} W_{f} & = & ∆ K E + ∆ P E \\ = & \frac{1}{2} m (v^{2} - u^{2}) + m g (h_{f} - h_{i}) \\ = & m [\frac{(v^{2} - u^{2})}{2} + g (h_{f} - h_{i})] \end{array}$

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