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Answer :
To find the work done on the object, we can use the work-energy theorem, which tells us that the work done is equal to the change in kinetic energy. The kinetic energy ([tex]$KE$[/tex]) of an object is given by
[tex]$$
KE = \frac{1}{2} m v^2,
$$[/tex]
where [tex]$m$[/tex] is the mass and [tex]$v$[/tex] is the velocity.
1. First, calculate the initial kinetic energy when the object is moving at [tex]$v_{\text{initial}} = 8.00\,\text{m/s}$[/tex]:
[tex]$$
KE_{\text{initial}} = \frac{1}{2} (60.0\,\text{kg}) (8.00\,\text{m/s})^2.
$$[/tex]
Simplifying,
[tex]$$
KE_{\text{initial}} = \frac{1}{2} (60.0)(64.00) = 1920.0\,\text{J}.
$$[/tex]
2. Next, calculate the final kinetic energy when the object is moving at [tex]$v_{\text{final}} = 4.00\,\text{m/s}$[/tex]:
[tex]$$
KE_{\text{final}} = \frac{1}{2} (60.0\,\text{kg}) (4.00\,\text{m/s})^2.
$$[/tex]
Simplifying,
[tex]$$
KE_{\text{final}} = \frac{1}{2} (60.0)(16.00) = 480.0\,\text{J}.
$$[/tex]
3. The work done ([tex]$W$[/tex]) on the object is the change in kinetic energy:
[tex]$$
W = KE_{\text{final}} - KE_{\text{initial}}.
$$[/tex]
Substituting the values we found:
[tex]$$
W = 480.0\,\text{J} - 1920.0\,\text{J} = -1440.0\,\text{J}.
$$[/tex]
The negative sign indicates that the work is done against the direction of the object's motion (i.e., energy is taken away from the object).
Thus, the work done on the object is
[tex]$$
\boxed{-1440\,\text{J}}.
$$[/tex]
[tex]$$
KE = \frac{1}{2} m v^2,
$$[/tex]
where [tex]$m$[/tex] is the mass and [tex]$v$[/tex] is the velocity.
1. First, calculate the initial kinetic energy when the object is moving at [tex]$v_{\text{initial}} = 8.00\,\text{m/s}$[/tex]:
[tex]$$
KE_{\text{initial}} = \frac{1}{2} (60.0\,\text{kg}) (8.00\,\text{m/s})^2.
$$[/tex]
Simplifying,
[tex]$$
KE_{\text{initial}} = \frac{1}{2} (60.0)(64.00) = 1920.0\,\text{J}.
$$[/tex]
2. Next, calculate the final kinetic energy when the object is moving at [tex]$v_{\text{final}} = 4.00\,\text{m/s}$[/tex]:
[tex]$$
KE_{\text{final}} = \frac{1}{2} (60.0\,\text{kg}) (4.00\,\text{m/s})^2.
$$[/tex]
Simplifying,
[tex]$$
KE_{\text{final}} = \frac{1}{2} (60.0)(16.00) = 480.0\,\text{J}.
$$[/tex]
3. The work done ([tex]$W$[/tex]) on the object is the change in kinetic energy:
[tex]$$
W = KE_{\text{final}} - KE_{\text{initial}}.
$$[/tex]
Substituting the values we found:
[tex]$$
W = 480.0\,\text{J} - 1920.0\,\text{J} = -1440.0\,\text{J}.
$$[/tex]
The negative sign indicates that the work is done against the direction of the object's motion (i.e., energy is taken away from the object).
Thus, the work done on the object is
[tex]$$
\boxed{-1440\,\text{J}}.
$$[/tex]
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