Question

A force F (vector)=(6i-2j)N acts on a particle undergoes a displacement delta r (Vector)=(3i+J)m. Find (a) the work done by the force on the particle and (b) the angle between F vector and r vector.

Answer 1

a) The work is given by the following dot product:

$$A = \mathbf{F}\cdot \mathbf{r}$$

where $\mathbf{F} = 6i - 2j$ and $\mathbf{r} = 3i + j$. Thus, obtain:

$$A = \mathbf{F}\cdot \mathbf{r} = (6i - 2j)\cdot (3i + j) = (6\cdot 3) + (-2)\cdot 1 = 16J$$

b) The angle between two vectors is:

$$\cos(\hat{\mathbf{F}\mathbf{r}}) = \dfrac{\mathbf{F}\cdot \mathbf{r}}{F\cdot r}$$

where $F = \sqrt{\mathbf{F}\cdot \mathbf{F}} = \sqrt{6^2 + 2^2} = 2\sqrt{10}$ is the lenght of vector $\mathbf{F}$ and $r = \sqrt{\mathbf{r}\cdot \mathbf{r}} = \sqrt{3^2 + 1^2} = \sqrt{10}$ is the length of vector $\mathbf{r}$. Thus, obtain:

$$\cos(\hat{\mathbf{F}\mathbf{r}}) = \dfrac{\mathbf{F}\cdot \mathbf{r}}{F\cdot r} = \dfrac{16}{2\sqrt{10}\cdot\sqrt{10}} = 0.8$$

The angle is:

$$\alpha = \arccos(0.8) \approx 36.9\degree$$

Answer. a) $16J$, b) $36.9\degree$.