Question

A constant force a={2,8} moves an object along a straight line from the point (3,6) to the point (10,14). Find the work done if the distance is measured in feet and the force is measured in pounds.

Answer 1

The work done by the constant force a={2,8} in moving the object from (3,6) to (10,14) is 238 foot-pounds.

Step-by-step explanation:

To find the work done by a constant force, we can use the equation W = |F| × |d| × cosθ, where W is the work done, F is the force, d is the displacement, and θ is the angle between the force and displacement vectors. In this case, the force is given as a={2,8}, and the displacement is from (3,6) to (10,14). To calculate the work, we need to find the magnitude of the force, the magnitude of the displacement, and the angle between them:

1. Magnitude of the force: |F| = √(2^2 + 8^2) = √(4 + 64) = √68 = 2√17

2. Magnitude of the displacement: |d| = √((10-3)^2 + (14-6)^2) = √(49 + 64) = √113

3. Angle between the force and displacement: cosθ = (2/|F|)×(7/|d|) = (2/(2√17))×(7/√113) = (7√17)/(2×√113)

Now, we can calculate the work:

W = |F| × |d| × cosθ

W = (2√17) × (√113) × ((7√17)/(2×√113)) = 2√17 × 7√17 = 14×17 = 238 foot-pounds (ft. lb)

Answer 2

The work done is 78 ft-lbs.

Step-by-step explanation:

The work W done by a force F in moving along a vector D is

`\textbf{W}= \textbf{F}\cdot  \textbf{D}`

In our case the displacement vector is

`\textbf{D}=<10-3,14-6>=<7,8>`

So the work done is

`\textbf{W}= \textbf{F}\cdot  \textbf{D}\\\textbf{W}=<2,8>\cdot <7,8>\\\textbf{W}=2\cdot7+8\cdot 8\\\textbf{W}=78`

We know that the unit of force is pounds and the distance is mesuared in feet, then the work done is 78 ft-lbs.