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9. A block with mass 0.500 kg is released from rest at the top of a vertical semicircular frictionless track that has radius 2.00 m. What is the normal force that the track exerts on the block as the block slides though the bottom of the track? 0.5ko (a) 4.9 N (6) 9.8 N T 14.7 la) 20.0 N le none of these answers botto mof track 2.0m

9. A block with mass 0.500 kg is released from rest at the top of a vertical semicircular frictionless track that has radius 2.00 m. What is the normal force that the track exerts on the block as the block slides though the bottom of the track? 0.5ko (a) 4.9 N (6) 9.8 N T 14.7 la) 20.0 N le none of these answers botto mof track 2.0m

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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**9. Problem Statement:** A block with a mass of 0.500 kg is released from rest at the top of a vertical semicircular frictionless track that has a radius of 2.00 m. What is the normal force that the track exerts on the block as the block slides through the bottom of the track? - (a) 4.9 N - (b) 9.8 N - (c) 14.7 N - (d) 20.0 N - (e) None of these answers ### Diagram Explanation: The diagram accompanying the problem shows: 1. A semicircular track with a radius of 2.0 meters. 2. A block labeled "0.5 kg" positioned at the top of the track. 3. The point at the bottom of the track is labeled as the "bottom of track." 4. An arrow indicating the direction of motion of the block from the top, down the semicircular path to the bottom. 5. The semicircular path has an arc with labeled radius to show the distance the block travels. ### Solution: To find the normal force at the bottom of the track: 1. **Determine Gravitational Potential Energy (GPE) at the top of the track:** - GPE = mgh = 0.5 kg * 9.8 m/s² * 2.0 m = 9.8 J. 2. **Convert GPE to Kinetic Energy (KE) at the bottom of the track, assuming no friction:** - KE = GPE - \( KE = \frac{1}{2} mv^2 \) - 9.8 J = \( \frac{1}{2} \) * 0.5 kg * v² - v² = 19.6 - v = √19.6 = 4.43 m/s 3. **Calculate the centripetal force needed to keep the block moving in a circle at the bottom of the track:** - Centripetal Force \( F_c \) = \( \frac{mv^2}{r} \) = \( \frac{0.5 kg * 4.43 m/s²}{2.0 m} \) = 4.9 N 4. **Determine the normal force:** - At the bottom of the
Transcribed Image Text:**9. Problem Statement:** A block with a mass of 0.500 kg is released from rest at the top of a vertical semicircular frictionless track that has a radius of 2.00 m. What is the normal force that the track exerts on the block as the block slides through the bottom of the track? - (a) 4.9 N - (b) 9.8 N - (c) 14.7 N - (d) 20.0 N - (e) None of these answers ### Diagram Explanation: The diagram accompanying the problem shows: 1. A semicircular track with a radius of 2.0 meters. 2. A block labeled "0.5 kg" positioned at the top of the track. 3. The point at the bottom of the track is labeled as the "bottom of track." 4. An arrow indicating the direction of motion of the block from the top, down the semicircular path to the bottom. 5. The semicircular path has an arc with labeled radius to show the distance the block travels. ### Solution: To find the normal force at the bottom of the track: 1. **Determine Gravitational Potential Energy (GPE) at the top of the track:** - GPE = mgh = 0.5 kg * 9.8 m/s² * 2.0 m = 9.8 J. 2. **Convert GPE to Kinetic Energy (KE) at the bottom of the track, assuming no friction:** - KE = GPE - \( KE = \frac{1}{2} mv^2 \) - 9.8 J = \( \frac{1}{2} \) * 0.5 kg * v² - v² = 19.6 - v = √19.6 = 4.43 m/s 3. **Calculate the centripetal force needed to keep the block moving in a circle at the bottom of the track:** - Centripetal Force \( F_c \) = \( \frac{mv^2}{r} \) = \( \frac{0.5 kg * 4.43 m/s²}{2.0 m} \) = 4.9 N 4. **Determine the normal force:** - At the bottom of the
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