Suppose that in each of the diagrams below, the force is applied to a block of mass 5.31 kg at rest on a level, frictionless surface. Calculate the block's speed in each is done. (a) F(N) 20.0 10.0 x (m) 8.00 4.00 m/s (b) F(N) 20.0 10.0 x (m) 8.00 4.00 m/s (c) F(N) 20.0 10.0 4.00 m/s
Suppose that in each of the diagrams below, the force is applied to a block of mass 5.31 kg at rest on a level, frictionless surface. Calculate the block's speed in each is done. (a) F(N) 20.0 10.0 x (m) 8.00 4.00 m/s (b) F(N) 20.0 10.0 x (m) 8.00 4.00 m/s (c) F(N) 20.0 10.0 4.00 m/s
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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![**Problem Statement**
Suppose that in each of the diagrams below, the force is applied to a block of mass 5.3 kg at rest on a level, frictionless surface. Calculate the block's speed in each case after the work is done.
**Diagrams and Analysis**
Each of the diagrams represents a force \( F \) as a function of position \( x \).
**(a) Diagram**
- The graph shows a constant force of 10.0 N applied from \( x = 0 \) to \( x = 4.00 \) m, and then the force is reduced to 0 N from \( x = 4.00 \) to \( x = 8.00 \) m.
**(b) Diagram**
- The graph shows a linearly increasing force from 0 N to 20.0 N over the distance from \( x = 0 \) to \( x = 4.00 \) m. The force remains at 20.0 N from \( x = 4.00 \) to \( x = 8.00 \) m.
**(c) Diagram**
- The force starts at 20.0 N from \( x = 0 \) to \( x = 4.00 \) m. From \( x = 4.00 \) to \( x = 8.00 \) m, the force decreases linearly back to 0 N.
**Tasks**
For each diagram, calculate the block's velocity at \( x = 8.00 \) m using the work-energy principle.
\[ \text{Work done} = \int F(x) \, dx = \Delta K \]
\[ \Delta K = \frac{1}{2} m v^2 - \frac{1}{2} m u^2 \]
Where \( m \) is the mass of the block, \( u \) is the initial velocity (0 m/s), and \( v \) is the final velocity. Calculate the speed of the block using the work done by the force.
**Boxes Provided**
- *Diagram (a):* _____ m/s
- *Diagram (b):* _____ m/s
- *Diagram (c):* _____ m/s](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F11ff91fa-fde1-47cc-bcac-170d9b2fe11c%2Faae81398-8415-4c63-a778-1631e1c5f080%2Ffevczdp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement**
Suppose that in each of the diagrams below, the force is applied to a block of mass 5.3 kg at rest on a level, frictionless surface. Calculate the block's speed in each case after the work is done.
**Diagrams and Analysis**
Each of the diagrams represents a force \( F \) as a function of position \( x \).
**(a) Diagram**
- The graph shows a constant force of 10.0 N applied from \( x = 0 \) to \( x = 4.00 \) m, and then the force is reduced to 0 N from \( x = 4.00 \) to \( x = 8.00 \) m.
**(b) Diagram**
- The graph shows a linearly increasing force from 0 N to 20.0 N over the distance from \( x = 0 \) to \( x = 4.00 \) m. The force remains at 20.0 N from \( x = 4.00 \) to \( x = 8.00 \) m.
**(c) Diagram**
- The force starts at 20.0 N from \( x = 0 \) to \( x = 4.00 \) m. From \( x = 4.00 \) to \( x = 8.00 \) m, the force decreases linearly back to 0 N.
**Tasks**
For each diagram, calculate the block's velocity at \( x = 8.00 \) m using the work-energy principle.
\[ \text{Work done} = \int F(x) \, dx = \Delta K \]
\[ \Delta K = \frac{1}{2} m v^2 - \frac{1}{2} m u^2 \]
Where \( m \) is the mass of the block, \( u \) is the initial velocity (0 m/s), and \( v \) is the final velocity. Calculate the speed of the block using the work done by the force.
**Boxes Provided**
- *Diagram (a):* _____ m/s
- *Diagram (b):* _____ m/s
- *Diagram (c):* _____ m/s
Expert Solution
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Step 1
Write the mass of the block.
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