If the force acts on an object that is initially at rest with mass 1kg, what is the objects final velocity? Hint: Work done = Change in KE F, (N) 15 10- 5- 0 Soloct ongi 1 2 3 -x (m)

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Chapter1: Units, Trigonometry. And Vectors
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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:**

If the force acts on an object that is initially at rest with mass 1 kg, what is the object's final velocity?

Hint: Work done = Change in KE (Kinetic Energy)

**Graph Description:**

The graph plots force (\( F_x \)) in newtons (N) on the y-axis against displacement (\( x \)) in meters (m) on the x-axis. It features a red line that starts at 15 N on the y-axis and decreases linearly to 0 N at 3 meters on the x-axis.

- At \( x = 0 \) m, \( F_x = 15 \) N.
- At \( x = 1.5 \) m, the force reaches about 7.5 N.
- At \( x = 3 \) m, \( F_x = 0 \).

**Instructions for Calculation:**

1. **Area under the curve:** Calculate the work done by finding the area under the force-displacement graph. This area represents the work done and can be calculated as the area of a triangle.
2. **Calculate the change in kinetic energy (KE):** Use the work-energy principle where work done = change in KE.
3. **Determine the final velocity:** Use the formula for kinetic energy, \( KE = \frac{1}{2} m v^2 \), to solve for the final velocity.

**Concepts:**

- **Work-Energy Principle**: The work done on an object is equal to its change in kinetic energy.
- **Kinetic Energy Formula**: \( KE = \frac{1}{2} m v^2 \)
- **Area of a Triangle**: \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \)

This problem is an application of physics concepts related to force, work, and energy, essential for understanding motion and dynamics.
Transcribed Image Text:**Problem Statement:** If the force acts on an object that is initially at rest with mass 1 kg, what is the object's final velocity? Hint: Work done = Change in KE (Kinetic Energy) **Graph Description:** The graph plots force (\( F_x \)) in newtons (N) on the y-axis against displacement (\( x \)) in meters (m) on the x-axis. It features a red line that starts at 15 N on the y-axis and decreases linearly to 0 N at 3 meters on the x-axis. - At \( x = 0 \) m, \( F_x = 15 \) N. - At \( x = 1.5 \) m, the force reaches about 7.5 N. - At \( x = 3 \) m, \( F_x = 0 \). **Instructions for Calculation:** 1. **Area under the curve:** Calculate the work done by finding the area under the force-displacement graph. This area represents the work done and can be calculated as the area of a triangle. 2. **Calculate the change in kinetic energy (KE):** Use the work-energy principle where work done = change in KE. 3. **Determine the final velocity:** Use the formula for kinetic energy, \( KE = \frac{1}{2} m v^2 \), to solve for the final velocity. **Concepts:** - **Work-Energy Principle**: The work done on an object is equal to its change in kinetic energy. - **Kinetic Energy Formula**: \( KE = \frac{1}{2} m v^2 \) - **Area of a Triangle**: \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \) This problem is an application of physics concepts related to force, work, and energy, essential for understanding motion and dynamics.
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