A frictionless plane inclined at 21° above the floor has length 3.3 m. A block starts from rest at the top of the inclined plane and slides down to the floor. What is its speed when it reaches the floor? A. 4.8 m/s B. 3.4 m/s C. 5.1 m/s D. cannot be determined with the information given
A frictionless plane inclined at 21° above the floor has length 3.3 m. A block starts from rest at the top of the inclined plane and slides down to the floor. What is its speed when it reaches the floor? A. 4.8 m/s B. 3.4 m/s C. 5.1 m/s D. cannot be determined with the information given
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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Transcribed Image Text:**Problem Statement:**
A frictionless plane inclined at 21° above the floor has a length of 3.3 meters. A block starts from rest at the top of the inclined plane and slides down to the floor. What is its speed when it reaches the floor?
**Options:**
- A. 4.8 m/s
- B. 3.4 m/s
- C. 5.1 m/s
- D. Cannot be determined with the information given
**Explanation:**
To solve this problem, one would typically use principles from physics, particularly energy conservation. With a frictionless incline, potential energy at the top is converted into kinetic energy at the bottom. The calculation would involve:
- Determining the height (h) using trigonometry: \( h = 3.3 \, \text{m} \times \sin(21^\circ) \).
- Using energy conservation: \( mgh = \frac{1}{2}mv^2 \) where m is mass, g is gravitational acceleration (9.8 m/s²), and v is velocity.
- Solving for v: \( v = \sqrt{2gh} \).
Note that the answer choice “D” is incorrect unless key information is missing, such as the gravitational constant, which is typically assumed to be 9.8 m/s² unless stated otherwise.
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