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Answer :
To find out how much work was done on the object, we will calculate the change in its kinetic energy.
The formula for kinetic energy (KE) is:
[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]
where:
- [tex]\( m \)[/tex] is the mass of the object,
- [tex]\( v \)[/tex] is the velocity of the object.
First, we calculate the initial kinetic energy when the object is moving at 8.00 m/s:
1. Initial kinetic energy:
[tex]\[ KE_{\text{initial}} = \frac{1}{2} \times 60.0 \, \text{kg} \times (8.00 \, \text{m/s})^2 \][/tex]
[tex]\[ KE_{\text{initial}} = \frac{1}{2} \times 60.0 \times 64 \][/tex]
[tex]\[ KE_{\text{initial}} = 1920.0 \, \text{Joules} \][/tex]
2. Final kinetic energy:
Next, calculate the kinetic energy after the object slows down to 4.00 m/s:
[tex]\[ KE_{\text{final}} = \frac{1}{2} \times 60.0 \, \text{kg} \times (4.00 \, \text{m/s})^2 \][/tex]
[tex]\[ KE_{\text{final}} = \frac{1}{2} \times 60.0 \times 16 \][/tex]
[tex]\[ KE_{\text{final}} = 480.0 \, \text{Joules} \][/tex]
3. Work done:
The work done on the object is the change in kinetic energy:
[tex]\[ \text{Work} = KE_{\text{final}} - KE_{\text{initial}} \][/tex]
[tex]\[ \text{Work} = 480.0 \, \text{J} - 1920.0 \, \text{J} \][/tex]
[tex]\[ \text{Work} = -1440.0 \, \text{Joules} \][/tex]
Therefore, the work done on the object is [tex]\(-1440 \, \text{J}\)[/tex], which indicates that energy was taken out of the system, slowing the object down. The correct choice is [tex]\(-1,440 \, \text{J}\)[/tex].
The formula for kinetic energy (KE) is:
[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]
where:
- [tex]\( m \)[/tex] is the mass of the object,
- [tex]\( v \)[/tex] is the velocity of the object.
First, we calculate the initial kinetic energy when the object is moving at 8.00 m/s:
1. Initial kinetic energy:
[tex]\[ KE_{\text{initial}} = \frac{1}{2} \times 60.0 \, \text{kg} \times (8.00 \, \text{m/s})^2 \][/tex]
[tex]\[ KE_{\text{initial}} = \frac{1}{2} \times 60.0 \times 64 \][/tex]
[tex]\[ KE_{\text{initial}} = 1920.0 \, \text{Joules} \][/tex]
2. Final kinetic energy:
Next, calculate the kinetic energy after the object slows down to 4.00 m/s:
[tex]\[ KE_{\text{final}} = \frac{1}{2} \times 60.0 \, \text{kg} \times (4.00 \, \text{m/s})^2 \][/tex]
[tex]\[ KE_{\text{final}} = \frac{1}{2} \times 60.0 \times 16 \][/tex]
[tex]\[ KE_{\text{final}} = 480.0 \, \text{Joules} \][/tex]
3. Work done:
The work done on the object is the change in kinetic energy:
[tex]\[ \text{Work} = KE_{\text{final}} - KE_{\text{initial}} \][/tex]
[tex]\[ \text{Work} = 480.0 \, \text{J} - 1920.0 \, \text{J} \][/tex]
[tex]\[ \text{Work} = -1440.0 \, \text{Joules} \][/tex]
Therefore, the work done on the object is [tex]\(-1440 \, \text{J}\)[/tex], which indicates that energy was taken out of the system, slowing the object down. The correct choice is [tex]\(-1,440 \, \text{J}\)[/tex].
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