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### Given Data:
- Mass of the body, [tex]\( m = 3.9 \)[/tex] kg
- Initial velocity at [tex]\( x = 1.6 \)[/tex] m, [tex]\( v_i = 10 \)[/tex] m/s
- Force component, [tex]\( F_x = -9x \)[/tex] N
- Initial position, [tex]\( x_i = 1.6 \)[/tex] m
### Part (a): Finding the velocity at [tex]\( x = 4.5 \)[/tex] m
1. Work-Energy Theorem: The work done by the force on the body as it moves from [tex]\( x_i \)[/tex] to [tex]\( x_f \)[/tex] equals the change in kinetic energy of the body.
2. Expression for Work Done: The work done by the force over a distance can be calculated using the formula:
[tex]\[ W = \int_{x_i}^{x_f} F_x \, dx \][/tex]
Given [tex]\( F_x = -9x \)[/tex], the work done [tex]\( W \)[/tex] from [tex]\( x_i = 1.6 \)[/tex] m to [tex]\( x_f = 4.5 \)[/tex] m is:
[tex]\[ W = \int_{1.6}^{4.5} (-9x) \, dx = -9 \int_{1.6}^{4.5} x \, dx \][/tex]
3. Solving the Integral:
[tex]\[ W = -9 \left[ \frac{x^2}{2} \right]_{1.6}^{4.5} = -9 \left( \frac{4.5^2}{2} - \frac{1.6^2}{2} \right) = -9 \left( \frac{20.25}{2} - \frac{2.56}{2} \right) = -9 \left( 10.125 - 1.28 \right) = -9 \times 8.845 = -79.605 \, \text{J} \][/tex]
4. Change in Kinetic Energy: According to the work-energy theorem:
[tex]\[ W = \Delta K = K_f - K_i \][/tex]
Initial kinetic energy [tex]\( K_i \)[/tex] is:
[tex]\[ K_i = \frac{1}{2} m v_i^2 = \frac{1}{2} \times 3.9 \times 10^2 = 195 \, \text{J} \][/tex]
Final kinetic energy [tex]\( K_f \)[/tex] is:
[tex]\[ K_f = K_i + W = 195 + (-79.605) = 115.395 \, \text{J} \][/tex]
5. Solving for Final Velocity:
[tex]\[ K_f = \frac{1}{2} m v_f^2 = 115.395 \][/tex]
[tex]\[ \frac{1}{2} \times 3.9 \times v_f^2 = 115.395 \][/tex]
[tex]\[ v_f^2 = \frac{115.395 \times 2}{3.9} = 59.1769 \][/tex]
[tex]\[ v_f = \sqrt{59.1769} \approx 7.69 \, \text{m/s} \][/tex]
So, the velocity of the body at [tex]\( x = 4.5 \)[/tex] m is approximately [tex]\( 7.69 \)[/tex] m/s.
### Part (b): Finding the position where the velocity is [tex]\( 3.1 \)[/tex] m/s
1. Final Kinetic Energy with [tex]\( v_f = 3.1 \)[/tex] m/s:
[tex]\[ K_f = \frac{1}{2} m v_f^2 = \frac{1}{2} \times 3.9 \times 3.1^2 = \frac{1}{2} \times 3.9 \times 9.61 = 18.7395 \, \text{J} \][/tex]
2. Work Done from [tex]\( x_i \)[/tex] to [tex]\( x_f \)[/tex]:
[tex]\[ W = \Delta K = K_f - K_i = 18.7395 - 195 = -176.2605 \, \text{J} \][/tex]
3. Solving for [tex]\( x \)[/tex]:
[tex]\[ W = \int_{1.6}^{x} (-9x) \, dx = -9 \left( \frac{x^2}{2} - \frac{1.6^2}{2} \right) \][/tex]
[tex]\[ -176.2605 = -9 \left( \frac{x^2}{2} - \frac{1.6^2}{2} \right) \][/tex]
Simplifying:
[tex]\[ -176.2605 = -4.5 \left( x^2 - 2.56 \right) \][/tex]
[tex]\[ 176.2605 = 4.5x^2 - 11.52 \][/tex]
[tex]\[ 187.7805 = 4.5x^2 \][/tex]
[tex]\[ x^2 = \frac{187.7805}{4.5} \approx 41.73 \][/tex]
[tex]\[ x = \sqrt{41.73} \approx 6.46 \, \text{m} \][/tex]
So, the position at which the body will have a velocity of [tex]\( 3.1 \)[/tex] m/s is approximately [tex]\( 6.46 \)[/tex] meters.
### Given Data:
- Mass of the body, [tex]\( m = 3.9 \)[/tex] kg
- Initial velocity at [tex]\( x = 1.6 \)[/tex] m, [tex]\( v_i = 10 \)[/tex] m/s
- Force component, [tex]\( F_x = -9x \)[/tex] N
- Initial position, [tex]\( x_i = 1.6 \)[/tex] m
### Part (a): Finding the velocity at [tex]\( x = 4.5 \)[/tex] m
1. Work-Energy Theorem: The work done by the force on the body as it moves from [tex]\( x_i \)[/tex] to [tex]\( x_f \)[/tex] equals the change in kinetic energy of the body.
2. Expression for Work Done: The work done by the force over a distance can be calculated using the formula:
[tex]\[ W = \int_{x_i}^{x_f} F_x \, dx \][/tex]
Given [tex]\( F_x = -9x \)[/tex], the work done [tex]\( W \)[/tex] from [tex]\( x_i = 1.6 \)[/tex] m to [tex]\( x_f = 4.5 \)[/tex] m is:
[tex]\[ W = \int_{1.6}^{4.5} (-9x) \, dx = -9 \int_{1.6}^{4.5} x \, dx \][/tex]
3. Solving the Integral:
[tex]\[ W = -9 \left[ \frac{x^2}{2} \right]_{1.6}^{4.5} = -9 \left( \frac{4.5^2}{2} - \frac{1.6^2}{2} \right) = -9 \left( \frac{20.25}{2} - \frac{2.56}{2} \right) = -9 \left( 10.125 - 1.28 \right) = -9 \times 8.845 = -79.605 \, \text{J} \][/tex]
4. Change in Kinetic Energy: According to the work-energy theorem:
[tex]\[ W = \Delta K = K_f - K_i \][/tex]
Initial kinetic energy [tex]\( K_i \)[/tex] is:
[tex]\[ K_i = \frac{1}{2} m v_i^2 = \frac{1}{2} \times 3.9 \times 10^2 = 195 \, \text{J} \][/tex]
Final kinetic energy [tex]\( K_f \)[/tex] is:
[tex]\[ K_f = K_i + W = 195 + (-79.605) = 115.395 \, \text{J} \][/tex]
5. Solving for Final Velocity:
[tex]\[ K_f = \frac{1}{2} m v_f^2 = 115.395 \][/tex]
[tex]\[ \frac{1}{2} \times 3.9 \times v_f^2 = 115.395 \][/tex]
[tex]\[ v_f^2 = \frac{115.395 \times 2}{3.9} = 59.1769 \][/tex]
[tex]\[ v_f = \sqrt{59.1769} \approx 7.69 \, \text{m/s} \][/tex]
So, the velocity of the body at [tex]\( x = 4.5 \)[/tex] m is approximately [tex]\( 7.69 \)[/tex] m/s.
### Part (b): Finding the position where the velocity is [tex]\( 3.1 \)[/tex] m/s
1. Final Kinetic Energy with [tex]\( v_f = 3.1 \)[/tex] m/s:
[tex]\[ K_f = \frac{1}{2} m v_f^2 = \frac{1}{2} \times 3.9 \times 3.1^2 = \frac{1}{2} \times 3.9 \times 9.61 = 18.7395 \, \text{J} \][/tex]
2. Work Done from [tex]\( x_i \)[/tex] to [tex]\( x_f \)[/tex]:
[tex]\[ W = \Delta K = K_f - K_i = 18.7395 - 195 = -176.2605 \, \text{J} \][/tex]
3. Solving for [tex]\( x \)[/tex]:
[tex]\[ W = \int_{1.6}^{x} (-9x) \, dx = -9 \left( \frac{x^2}{2} - \frac{1.6^2}{2} \right) \][/tex]
[tex]\[ -176.2605 = -9 \left( \frac{x^2}{2} - \frac{1.6^2}{2} \right) \][/tex]
Simplifying:
[tex]\[ -176.2605 = -4.5 \left( x^2 - 2.56 \right) \][/tex]
[tex]\[ 176.2605 = 4.5x^2 - 11.52 \][/tex]
[tex]\[ 187.7805 = 4.5x^2 \][/tex]
[tex]\[ x^2 = \frac{187.7805}{4.5} \approx 41.73 \][/tex]
[tex]\[ x = \sqrt{41.73} \approx 6.46 \, \text{m} \][/tex]
So, the position at which the body will have a velocity of [tex]\( 3.1 \)[/tex] m/s is approximately [tex]\( 6.46 \)[/tex] meters.
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