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To determine which of the given objects has the highest kinetic energy, we need to calculate the kinetic energy for each object individually. Here’s a detailed step-by-step solution:
1. Cart:
- A 1 kg cart moving at 1 m/s.
- Kinetic energy (KE) is given by the formula:
[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]
- Substituting the values:
[tex]\[ KE = \frac{1}{2} \times 1 \text{ kg} \times (1 \text{ m/s})^2 = 0.5 \text{ J} \][/tex]
2. Solid sphere:
- A 1 kg solid sphere with a radius of 1 m rolling along a track at 1 rad/s.
- The sphere has both translational and rotational kinetic energy.
- Translational kinetic energy:
[tex]\[ KE_{\text{trans}} = \frac{1}{2} m v^2 \][/tex]
Since [tex]\( v = \omega \times r \)[/tex], where [tex]\( \omega = 1 \text{ rad/s} \)[/tex] and [tex]\( r = 1 \text{ m} \)[/tex], we have:
[tex]\[ v = 1 \text{ m/s} \][/tex]
[tex]\[ KE_{\text{trans}} = \frac{1}{2} \times 1 \text{ kg} \times (1 \text{ m/s})^2 = 0.5 \text{ J} \][/tex]
- Rotational kinetic energy:
[tex]\[ KE_{\text{rot}} = \frac{1}{2} I \omega^2 \][/tex]
For a solid sphere, the moment of inertia [tex]\( I \)[/tex] is:
[tex]\[ I = \frac{2}{5} m r^2 = \frac{2}{5} \times 1 \text{ kg} \times (1 \text{ m})^2 = \frac{2}{5} \text{ kg} \cdot \text{m}^2 \][/tex]
[tex]\[ KE_{\text{rot}} = \frac{1}{2} \times \frac{2}{5} \text{ kg} \cdot \text{m}^2 \times (1 \text{ rad/s})^2 = \frac{1}{5} \text{ J} = 0.2 \text{ J} \][/tex]
- Total kinetic energy for the sphere:
[tex]\[ KE_{\text{total}} = KE_{\text{trans}} + KE_{\text{rot}} = 0.5 \text{ J} + 0.2 \text{ J} = 0.7 \text{ J} \][/tex]
3. Cylinder:
- A 1 kg cylinder with a radius of 1 m rolling along a track at 1 rad/s.
- The cylinder has both translational and rotational kinetic energy.
- Translational kinetic energy:
[tex]\[ KE_{\text{trans}} = \frac{1}{2} m v^2 \][/tex]
Since [tex]\( v = \omega \times r \)[/tex], where [tex]\( \omega = 1 \text{ rad/s} \)[/tex] and [tex]\( r = 1 \text{ m} \)[/tex], we have:
[tex]\[ v = 1 \text{ m/s} \][/tex]
[tex]\[ KE_{\text{trans}} = \frac{1}{2} \times 1 \text{ kg} \times (1 \text{ m/s})^2 = 0.5 \text{ J} \][/tex]
- Rotational kinetic energy:
[tex]\[ KE_{\text{rot}} = \frac{1}{2} I \omega^2 \][/tex]
For a cylinder (thin cylindrical shell), the moment of inertia [tex]\( I \)[/tex] is:
[tex]\[ I = m r^2 = 1 \text{ kg} \times (1 \text{ m})^2 = 1 \text{ kg} \cdot \text{m}^2 \][/tex]
[tex]\[ KE_{\text{rot}} = \frac{1}{2} \times 1 \text{ kg} \cdot \text{m}^2 \times (1 \text{ rad/s})^2 = 0.5 \text{ J} \][/tex]
- Total kinetic energy for the cylinder:
[tex]\[ KE_{\text{total}} = KE_{\text{trans}} + KE_{\text{rot}} = 0.5 \text{ J} + 0.5 \text{ J} = 1 \text{ J} \][/tex]
4. Stone:
- A 1 kg stone moving in a circle of radius 1 m at 1 rad/s.
- The kinetic energy for an object moving in a circle is purely translational:
[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]
Since [tex]\( v = \omega \times r \)[/tex], where [tex]\( \omega = 1 \text{ rad/s} \)[/tex] and [tex]\( r = 1 \text{ m} \)[/tex], we have:
[tex]\[ v = 1 \text{ m/s} \][/tex]
[tex]\[ KE = \frac{1}{2} \times 1 \text{ kg} \times (1 \text{ m/s})^2 = 0.5 \text{ J} \][/tex]
By comparing the kinetic energies calculated for each object:
- Cart: 0.5 J
- Solid sphere: 0.7 J
- Cylinder: 1.0 J
- Stone: 0.5 J
The cylinder, with a kinetic energy of 1.0 J, has the highest kinetic energy among the objects listed.
1. Cart:
- A 1 kg cart moving at 1 m/s.
- Kinetic energy (KE) is given by the formula:
[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]
- Substituting the values:
[tex]\[ KE = \frac{1}{2} \times 1 \text{ kg} \times (1 \text{ m/s})^2 = 0.5 \text{ J} \][/tex]
2. Solid sphere:
- A 1 kg solid sphere with a radius of 1 m rolling along a track at 1 rad/s.
- The sphere has both translational and rotational kinetic energy.
- Translational kinetic energy:
[tex]\[ KE_{\text{trans}} = \frac{1}{2} m v^2 \][/tex]
Since [tex]\( v = \omega \times r \)[/tex], where [tex]\( \omega = 1 \text{ rad/s} \)[/tex] and [tex]\( r = 1 \text{ m} \)[/tex], we have:
[tex]\[ v = 1 \text{ m/s} \][/tex]
[tex]\[ KE_{\text{trans}} = \frac{1}{2} \times 1 \text{ kg} \times (1 \text{ m/s})^2 = 0.5 \text{ J} \][/tex]
- Rotational kinetic energy:
[tex]\[ KE_{\text{rot}} = \frac{1}{2} I \omega^2 \][/tex]
For a solid sphere, the moment of inertia [tex]\( I \)[/tex] is:
[tex]\[ I = \frac{2}{5} m r^2 = \frac{2}{5} \times 1 \text{ kg} \times (1 \text{ m})^2 = \frac{2}{5} \text{ kg} \cdot \text{m}^2 \][/tex]
[tex]\[ KE_{\text{rot}} = \frac{1}{2} \times \frac{2}{5} \text{ kg} \cdot \text{m}^2 \times (1 \text{ rad/s})^2 = \frac{1}{5} \text{ J} = 0.2 \text{ J} \][/tex]
- Total kinetic energy for the sphere:
[tex]\[ KE_{\text{total}} = KE_{\text{trans}} + KE_{\text{rot}} = 0.5 \text{ J} + 0.2 \text{ J} = 0.7 \text{ J} \][/tex]
3. Cylinder:
- A 1 kg cylinder with a radius of 1 m rolling along a track at 1 rad/s.
- The cylinder has both translational and rotational kinetic energy.
- Translational kinetic energy:
[tex]\[ KE_{\text{trans}} = \frac{1}{2} m v^2 \][/tex]
Since [tex]\( v = \omega \times r \)[/tex], where [tex]\( \omega = 1 \text{ rad/s} \)[/tex] and [tex]\( r = 1 \text{ m} \)[/tex], we have:
[tex]\[ v = 1 \text{ m/s} \][/tex]
[tex]\[ KE_{\text{trans}} = \frac{1}{2} \times 1 \text{ kg} \times (1 \text{ m/s})^2 = 0.5 \text{ J} \][/tex]
- Rotational kinetic energy:
[tex]\[ KE_{\text{rot}} = \frac{1}{2} I \omega^2 \][/tex]
For a cylinder (thin cylindrical shell), the moment of inertia [tex]\( I \)[/tex] is:
[tex]\[ I = m r^2 = 1 \text{ kg} \times (1 \text{ m})^2 = 1 \text{ kg} \cdot \text{m}^2 \][/tex]
[tex]\[ KE_{\text{rot}} = \frac{1}{2} \times 1 \text{ kg} \cdot \text{m}^2 \times (1 \text{ rad/s})^2 = 0.5 \text{ J} \][/tex]
- Total kinetic energy for the cylinder:
[tex]\[ KE_{\text{total}} = KE_{\text{trans}} + KE_{\text{rot}} = 0.5 \text{ J} + 0.5 \text{ J} = 1 \text{ J} \][/tex]
4. Stone:
- A 1 kg stone moving in a circle of radius 1 m at 1 rad/s.
- The kinetic energy for an object moving in a circle is purely translational:
[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]
Since [tex]\( v = \omega \times r \)[/tex], where [tex]\( \omega = 1 \text{ rad/s} \)[/tex] and [tex]\( r = 1 \text{ m} \)[/tex], we have:
[tex]\[ v = 1 \text{ m/s} \][/tex]
[tex]\[ KE = \frac{1}{2} \times 1 \text{ kg} \times (1 \text{ m/s})^2 = 0.5 \text{ J} \][/tex]
By comparing the kinetic energies calculated for each object:
- Cart: 0.5 J
- Solid sphere: 0.7 J
- Cylinder: 1.0 J
- Stone: 0.5 J
The cylinder, with a kinetic energy of 1.0 J, has the highest kinetic energy among the objects listed.
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