A machine carries a 4.0 kg package from an initial position of `vec(d)_(i)=(0.50 m)hat(i)+(0.75 m)hat(j)+(0.20 m)hat(k)` at `t=0` to a final position of `vec(d)_(f)=(7.50 m)hat(i)+(12.0 m) hat(k)` at `t=12` s. The constant force applied by the machine on the package is `vec(F)= (2.00 N) hat(i) + (4.00 N) hat(j) + (6.00 N) hat(k)`. For that displacement, find (a) the work done on the package by the machine's force and (b) the average power of the machine's force on the package.

Answer

To solve the problem step-by-step, we will find the work done on the package by the machine's force and then calculate the average power exerted by the machine on the package.

### Step 1: Calculate the Displacement Vector
The displacement vector \( \vec{d} \) is given by the difference between the final position vector \( \vec{d}_f \) and the initial position vector \( \vec{d}_i \).

\[ \vec{d}_i = (0.50 \, \hat{i} + 0.75 \, \hat{j} + 0.20 \, \hat{k}) \, \text{m} \]
\[ \vec{d}_f = (7.50 \, \hat{i} + 0 \, \hat{j} + 12.0 \, \hat{k}) \, \text{m} \]

Calculating the displacement:

\[ \vec{d} = \vec{d}_f - \vec{d}_i = (7.50 - 0.50) \hat{i} + (0 - 0.75) \hat{j} + (12.0 - 0.20) \hat{k} \]
\[ \vec{d} = 7.0 \, \hat{i} - 0.75 \, \hat{j} + 11.8 \, \hat{k} \, \text{m} \]

### Step 2: Calculate the Work Done by the Machine's Force
The work done \( W \) by a force \( \vec{F} \) on an object during a displacement \( \vec{d} \) is given by the dot product:

\[ W = \vec{F} \cdot \vec{d} \]

Given the force vector:

\[ \vec{F} = (2.00 \, \hat{i} + 4.00 \, \hat{j} + 6.00 \, \hat{k}) \, \text{N} \]

Calculating the dot product:

\[ W = (2.00 \, \hat{i} + 4.00 \, \hat{j} + 6.00 \, \hat{k}) \cdot (7.0 \, \hat{i} - 0.75 \, \hat{j} + 11.8 \, \hat{k}) \]

Calculating each component:

\[ W = (2.00 \times 7.0) + (4.00 \times -0.75) + (6.00 \times 11.8) \]
\[ W = 14.0 - 3.0 + 70.8 \]
\[ W = 81.8 \, \text{Joules} \]

### Step 3: Calculate the Average Power
The average power \( P \) is defined as the work done divided by the time taken:

\[ P = \frac{W}{t} \]

Given that the time \( t = 12 \, \text{s} \):

\[ P = \frac{81.8 \, \text{J}}{12 \, \text{s}} \approx 6.815 \, \text{W} \]

### Final Answers
(a) The work done on the package by the machine's force is \( 81.8 \, \text{J} \).

(b) The average power of the machine's force on the package is approximately \( 6.82 \, \text{W} \).

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