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

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Chapter1: Units, Trigonometry. And Vectors
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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
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
Step 1

Write the mass of the block.

m=5.31 kg

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