Question
A 5-lb. force acting in the direction of left angle bracket 5 comma negative 3 right angle bracket moves an object just over 12 ft. from point left parenthesis 0 comma 6 right parenthesis to left parenthesis 7 comma negative 4 right parenthesis. Find the work done to move the object to the nearest foot-pound.
(1 point)
Responses
11 ft.·lbs.
11 ft.·lbs.
34 ft.·lbs.
34 ft.·lbs.
56 ft.·lbs.
56 ft.·lbs.
61 ft.·lbs.
(1 point)
Responses
11 ft.·lbs.
11 ft.·lbs.
34 ft.·lbs.
34 ft.·lbs.
56 ft.·lbs.
56 ft.·lbs.
61 ft.·lbs.
Answers
GPT-4o mini
To calculate the work done on the object when it is moved, we can use the formula for work:
\[
W = F \cdot d \cdot \cos(\theta)
\]
where \(W\) is work, \(F\) is the magnitude of the force, \(d\) is the distance moved, and \(\theta\) is the angle between the force vector and the direction of movement.
### Step 1: Calculate the Distance Moved
The object moves from point \((0, 6)\) to \((7, -4)\). We can find the distance \(d\) using the distance formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
Plugging in the points:
\[
d = \sqrt{(7 - 0)^2 + (-4 - 6)^2} = \sqrt{7^2 + (-10)^2} = \sqrt{49 + 100} = \sqrt{149} \approx 12.21 \text{ ft}
\]
### Step 2: Determine the Direction of the Force Vector
The direction of the force vector is given as \(\langle 5, -3 \rangle\).
### Step 3: Calculate the Magnitude of the Force Vector
The magnitude of the direction vector is:
\[
\| \langle 5, -3 \rangle \| = \sqrt{5^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34}
\]
### Step 4: Find the Unit Vector of the Direction
To find the unit vector in the direction of the movement:
\[
\text{Unit vector} = \left\langle \frac{5}{\sqrt{34}}, \frac{-3}{\sqrt{34}} \right\rangle
\]
### Step 5: Calculate Work Done
The work done \(W\) can be calculated with:
\[
W = F \cdot d
\]
where \(F = 5 \text{ lbs}\). The force is acting in the direction of the unit vector we found, so we take the dot product as the full force \(F\) is acting along that vector.
Now we have:
\[
W = F \cdot d
\]
Substituting the values:
\[
W \approx 5 \text{ lbs} \times 12.21 \text{ ft} \approx 61.05 \text{ ft-lbs}
\]
### Final Result
Rounding to the nearest foot-pound gives us:
\[
\text{Work done} \approx 61 \text{ ft-lbs}
\]
Thus, the answer is **61 ft.·lbs.**
\[
W = F \cdot d \cdot \cos(\theta)
\]
where \(W\) is work, \(F\) is the magnitude of the force, \(d\) is the distance moved, and \(\theta\) is the angle between the force vector and the direction of movement.
### Step 1: Calculate the Distance Moved
The object moves from point \((0, 6)\) to \((7, -4)\). We can find the distance \(d\) using the distance formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
Plugging in the points:
\[
d = \sqrt{(7 - 0)^2 + (-4 - 6)^2} = \sqrt{7^2 + (-10)^2} = \sqrt{49 + 100} = \sqrt{149} \approx 12.21 \text{ ft}
\]
### Step 2: Determine the Direction of the Force Vector
The direction of the force vector is given as \(\langle 5, -3 \rangle\).
### Step 3: Calculate the Magnitude of the Force Vector
The magnitude of the direction vector is:
\[
\| \langle 5, -3 \rangle \| = \sqrt{5^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34}
\]
### Step 4: Find the Unit Vector of the Direction
To find the unit vector in the direction of the movement:
\[
\text{Unit vector} = \left\langle \frac{5}{\sqrt{34}}, \frac{-3}{\sqrt{34}} \right\rangle
\]
### Step 5: Calculate Work Done
The work done \(W\) can be calculated with:
\[
W = F \cdot d
\]
where \(F = 5 \text{ lbs}\). The force is acting in the direction of the unit vector we found, so we take the dot product as the full force \(F\) is acting along that vector.
Now we have:
\[
W = F \cdot d
\]
Substituting the values:
\[
W \approx 5 \text{ lbs} \times 12.21 \text{ ft} \approx 61.05 \text{ ft-lbs}
\]
### Final Result
Rounding to the nearest foot-pound gives us:
\[
\text{Work done} \approx 61 \text{ ft-lbs}
\]
Thus, the answer is **61 ft.·lbs.**