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Sagot :
To find the work done in moving the object, we will follow these steps:
1. Calculate the distance vector.
2. Normalize the direction vector.
3. Compute the projection of the distance vector onto the direction vector.
4. Calculate work done using the formula [tex]\( W = \mathbf{F} \cdot \mathbf{d} \)[/tex].
Let's go through each step in detail.
### Step 1: Calculate the Distance Vector
The object moves from point [tex]\((0, 6)\)[/tex] to point [tex]\((7, -4)\)[/tex]. The distance vector can be found by subtracting the initial point coordinates from the final point coordinates:
[tex]\[ \mathbf{d} = (7 - 0, -4 - 6) = (7, -10) \][/tex]
### Step 2: Normalize the Direction Vector
The given direction vector is [tex]\(\langle 5, -3 \rangle\)[/tex]. To normalize it, find its magnitude and divide each component by the magnitude.
Magnitude of the direction vector:
[tex]\[ \|\mathbf{v}\| = \sqrt{5^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34} \][/tex]
Normalized direction vector:
[tex]\[ \mathbf{v}_{\text{unit}} = \left( \frac{5}{\sqrt{34}}, \frac{-3}{\sqrt{34}} \right) \][/tex]
### Step 3: Compute the Projection
The projection of [tex]\(\mathbf{d}\)[/tex] onto the unit direction vector [tex]\(\mathbf{v}_{\text{unit}}\)[/tex] is computed using the dot product and the magnitude of [tex]\(\mathbf{d}\)[/tex].
[tex]\[ \mathbf{d} \cdot \mathbf{v}_{\text{unit}} = \left( 7, -10 \right) \cdot \left( \frac{5}{\sqrt{34}}, \frac{-3}{\sqrt{34}} \right) \][/tex]
[tex]\[ \mathbf{d} \cdot \mathbf{v}_{\text{unit}} = \frac{1}{\sqrt{34}} (7 \times 5 + (-10) \times (-3)) \][/tex]
[tex]\[ \mathbf{d} \cdot \mathbf{v}_{\text{unit}} = \frac{1}{\sqrt{34}} (35 + 30) = \frac{65}{\sqrt{34}} \][/tex]
### Step 4: Calculate Work Done
To find the work done, we multiply the force magnitude by the projection of the distance vector onto the direction vector.
Given the force magnitude is 5 lbs:
[tex]\[ W = 5 \times \frac{65}{\sqrt{34}} \][/tex]
Simplify the expression:
[tex]\[ W = \frac{5 \times 65}{\sqrt{34}} = \frac{325}{\sqrt{34}} \][/tex]
We can rationalize the denominator:
[tex]\[ W = \frac{325 \sqrt{34}}{34} = \frac{325 \times \sqrt{34}}{34} \approx 61 \][/tex]
Therefore, the work done in moving the object is [tex]\( \boxed{61} \)[/tex] ft-lbs.
1. Calculate the distance vector.
2. Normalize the direction vector.
3. Compute the projection of the distance vector onto the direction vector.
4. Calculate work done using the formula [tex]\( W = \mathbf{F} \cdot \mathbf{d} \)[/tex].
Let's go through each step in detail.
### Step 1: Calculate the Distance Vector
The object moves from point [tex]\((0, 6)\)[/tex] to point [tex]\((7, -4)\)[/tex]. The distance vector can be found by subtracting the initial point coordinates from the final point coordinates:
[tex]\[ \mathbf{d} = (7 - 0, -4 - 6) = (7, -10) \][/tex]
### Step 2: Normalize the Direction Vector
The given direction vector is [tex]\(\langle 5, -3 \rangle\)[/tex]. To normalize it, find its magnitude and divide each component by the magnitude.
Magnitude of the direction vector:
[tex]\[ \|\mathbf{v}\| = \sqrt{5^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34} \][/tex]
Normalized direction vector:
[tex]\[ \mathbf{v}_{\text{unit}} = \left( \frac{5}{\sqrt{34}}, \frac{-3}{\sqrt{34}} \right) \][/tex]
### Step 3: Compute the Projection
The projection of [tex]\(\mathbf{d}\)[/tex] onto the unit direction vector [tex]\(\mathbf{v}_{\text{unit}}\)[/tex] is computed using the dot product and the magnitude of [tex]\(\mathbf{d}\)[/tex].
[tex]\[ \mathbf{d} \cdot \mathbf{v}_{\text{unit}} = \left( 7, -10 \right) \cdot \left( \frac{5}{\sqrt{34}}, \frac{-3}{\sqrt{34}} \right) \][/tex]
[tex]\[ \mathbf{d} \cdot \mathbf{v}_{\text{unit}} = \frac{1}{\sqrt{34}} (7 \times 5 + (-10) \times (-3)) \][/tex]
[tex]\[ \mathbf{d} \cdot \mathbf{v}_{\text{unit}} = \frac{1}{\sqrt{34}} (35 + 30) = \frac{65}{\sqrt{34}} \][/tex]
### Step 4: Calculate Work Done
To find the work done, we multiply the force magnitude by the projection of the distance vector onto the direction vector.
Given the force magnitude is 5 lbs:
[tex]\[ W = 5 \times \frac{65}{\sqrt{34}} \][/tex]
Simplify the expression:
[tex]\[ W = \frac{5 \times 65}{\sqrt{34}} = \frac{325}{\sqrt{34}} \][/tex]
We can rationalize the denominator:
[tex]\[ W = \frac{325 \sqrt{34}}{34} = \frac{325 \times \sqrt{34}}{34} \approx 61 \][/tex]
Therefore, the work done in moving the object is [tex]\( \boxed{61} \)[/tex] ft-lbs.
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