1. The problem is to understand the concept of \textbf{work} in physics and calculate it. 2. \textbf{Work} is defined as the dot product of force and displacement vectors: $$ W = \vec{F} \cdot \vec{d} $$ 3. When force and displacement are in the same direction, this simplifies to: $$ W = F \times d $$ where $F$ is the magnitude of force and $d$ is the magnitude of displacement. 4. Work done is measured in joules (J) where $1\ \text{joule} = 1\ \text{newton} \times 1\ \text{meter}$. 5. If a force $F$ acts at an angle $\theta$ to the direction of displacement $d$, then: $$ W = Fd \cos \theta $$ 6. In summary, to calculate work you need: the magnitude of the force, displacement, and the angle between the two. 7. \textbf{Example}: If a force of 10 newtons moves an object 5 meters in the same direction, the work done is: $$ W = 10 \times 5 = 50 \text{ joules} $$ This concludes the explanation and computation of work.