### Understanding Work Done by Friction in a Circular TrackA small block with a mass of 1 kg slides in a vertical circle of radius 0.5 m on the inside of a circular track. There is friction between the track and the block. At the bottom of the block's path, the speed of the block is 5 m/s, and at the top of the block's path, the speed of the block is 1 m/s. The question is: "In going from the bottom to the top of the track, how much work is done by the friction force on the block?"#### Answer Choices:- O -11.3 J- O 10.0 J- O -9.90 J- O Cannot be calculated from the given information- O -2.20 J#### Detailed Explanation:To solve this problem, consider the following points:- **Kinetic Energy at the Bottom:** \( KE_{bottom} = \frac{1}{2} m v^2 = \frac{1}{2} \times 1 \times 5^2 = 12.5 \, \text{J} \)- **Kinetic Energy at the Top:** \( KE_{top} = \frac{1}{2} m v^2 = \frac{1}{2} \times 1 \times 1^2 = 0.5 \, \text{J} \)- **Potential Energy at the Bottom:** It's zero since we are taking the bottom as the reference point \( PE_{bottom} = 0 \).- **Potential Energy at the Top:** \( PE_{top} = mgh = 1 \times 9.8 \times 1 = 9.8 \, \text{J} \) (since the height difference \( h \) is 1 m, the diameter of the circle).The work-energy principle states that:\[ W_{\mathrm{friction}} = \Delta KE + \Delta PE \]Substituting the values:\[ W_{\mathrm{friction}} = (0.5 - 12.5) + (9.8 - 0) \]\[ W_{\mathrm{friction}} = -12 + 9.8 \]\[ W_{\mathrm{friction}} = -2.2 \, \text{J} \]So, the correct answer is:

Name: ### Understanding Work Done by Friction in a Circular TrackA small bl
Uploaded: 2024-03-20T06:58:04.000Z
Channel: Bartleby
Description: ### Understanding Work Done by Friction in a Circular TrackA small block with a mass of 1 kg slides in a vertical circle of radius 0.5 m on the inside of a circular track. There is friction between the track and the block. At the bottom of the block