Question

A force F = (2i + 9j + 5.3K) * N acts on a 2.90 kg object moves in time interval 2.10 s from an initial posi- F_{3} = (2.7 * i - 2.9 * j + 5.5K) m to a final position \ - 4.1 hat h +3.30 hat prime +5.40 hat h )n . Find (a) the work done on the object vec r_{2} = by the force in that time interval, (b) the average power due to the fored during that time interval, and (c) the angle between vectors 7 and 7

Answer 1

The work done is 7.45 Joules, (b) the average power is approximately 3.55 Watts, and (c) the angle between vectors is about 85.6 degrees.

(a) Work Done:

The work done
`(\(W\))` is given by the dot product of force
`(\(\mathbf{F}\))` and displacement
`(\(\mathbf{r}\))`:

`\[ W = \mathbf{F} \cdot \mathbf{r} \]`

Substitute the given values and calculate
`\(W\)`.

**(b) Average Power:**

The average power
`(\(P_{\text{avg}}\))` is the work done per unit time:

`\[ P_{\text{avg}} = (W)/(\Delta t) \]`

where
`\(\Delta t\)` is the time interval. Substitute
`\(W\)` and
`\(\Delta t\)` to find
`\(P_{\text{avg}}\)`.

(c) Angle between Vectors:

The angle
`(\(\theta\))` between two vectors is given by:

`\[ \cos(\theta) = \frac{\mathbf{F} \cdot \mathbf{r}}{|\mathbf{F}| \cdot |\mathbf{r}|} \]`

Calculate
`\(\theta\)` using the dot product formula.

(a) Work Done:

`\[ W = \mathbf{F} \cdot \mathbf{r} \]`

`\[ W = (2\hat{i} + 9\hat{j} + 5.3\hat{k}) \cdot (2.7\hat{i} - 2.9\hat{j} + 5.5\hat{k}) \]`

`\[ W = 2(2.7) + 9(-2.9) + 5.3(5.5) \]`

`\[ W = 5.4 - 26.1 + 28.15 \]`

`\[ W = 7.45 \, \text{Joules} \]`

(b) Average Power:

`\[ P_{\text{avg}} = (W)/(\Delta t) \]`

`\[ P_{\text{avg}} = (7.45)/(2.10) \]`

`\[ P_{\text{avg}} \approx 3.55 \, \text{Watts} \]`

(c) Angle between Vectors:

`\[ \cos(\theta) = \frac{\mathbf{F} \cdot \mathbf{r}}{|\mathbf{F}| \cdot |\mathbf{r}|} \]`

`\[ \cos(\theta) = (7.45)/(|(2i + 9j + 5.3k)| \cdot |(2.7i - 2.9j + 5.5k)|) \]`

`\[ \cos(\theta) \approx (7.45)/(√(110.89) \cdot √(82.29)) \]`

`\[ \cos(\theta) \approx (7.45)/(10.54 \cdot 9.07) \]`

`\[ \cos(\theta) \approx 0.081 \]`

`\[ \theta \approx \cos^(-1)(0.081) \]`

`\[ \theta \approx 85.6^\circ \]`

So, (a) the work done is 7.45 Joules, (b) the average power is approximately 3.55 Watts, and (c) the angle between vectors is about 85.6 degrees.