Find the work (in $\mathbf{j}$) done by a force $\mathbf{F} = 4\mathbf{i} - 6\mathbf{j} + 9\mathbf{k}$ that moves an object from the point $(0, 10, 4)$ to the point $(6, 16, 18)$ along a straight line. The distance is measured in meters and the force in newtons.

Neetesh Kumar | December 18, 2024

Calculus Homework Help

This is the solution to Math 1C
Assignment: 12.3 Question Number 18
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Step-by-step solution:

The formula for work done is:
$W = \mathbf{F} \cdot \mathbf{d}$,
where:

• $\mathbf{F}$ is the force vector,
• $\mathbf{d}$ is the displacement vector.

Step 1: Find the displacement vector $\mathbf{d}$:

The displacement vector $\mathbf{d}$ is calculated as:
$\mathbf{d} = \mathbf{r}_2 - \mathbf{r}_1$,
where $\mathbf{r}_1 = (0, 10, 4)$ and $\mathbf{r}_2 = (6, 16, 18)$ are the initial and final position vectors.

Subtract the components:
$\mathbf{d} = \langle 6 - 0, 16 - 10, 18 - 4 \rangle$.

Simplify:
$\mathbf{d} = \langle 6, 6, 14 \rangle$.

Step 2: Compute the dot product $\mathbf{F} \cdot \mathbf{d}$:

The force vector is $\mathbf{F} = \langle 4, -6, 9 \rangle$ and the displacement vector is $\mathbf{d} = \langle 6, 6, 14 \rangle$.
The dot product is:
$\mathbf{F} \cdot \mathbf{d} = F_1d_1 + F_2d_2 + F_3d_3$,
where $F_1, F_2, F_3$ are the components of $\mathbf{F}$, and $d_1, d_2, d_3$ are the components of $\mathbf{d}$.

Substitute the values:
$\mathbf{F} \cdot \mathbf{d} = (4)(6) + (-6)(6) + (9)(14)$.

Simplify each term:
$\mathbf{F} \cdot \mathbf{d} = 24 - 36 + 126$.

Combine:
$\mathbf{F} \cdot \mathbf{d} = 114$.

Final Answer:

The work done by the force is:

$W = \boxed{114 \ \text{J}}$

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