Problem 61 - Algebra and Trigonometry

Question

A force of 4 pounds acts in the direction of \(50^{\circ}\) to the horizontal. The force moves an object along a straight line from the point \((3,7)\) to the point \((8,10),\) with distance measured in feet. Find the work done by the force.

Step-by-Step Solution

Answer

The work done by the force is approximately \(19.28\) foot-pounds.

Step 1: Locate the displacement vector

The displacement vector can be computed by subtracting the coordinates of the initial point from the coordinates of the final point. Therefore, it's \( (8-3,10-7) \) which simplifies to \( (5, 3) \).

Step 2: Compute the force vector

The force vector will point in the direction of \( 50^{\circ} \) to the horizontal and have a magnitude of 4 pounds. Convert the angle to radians and decompose the force into its horizontal and vertical components to get the force vector: \( (4 \cos(50^\circ), 4 \sin(50^\circ)) \). In approximate decimal form this becomes: \( (2.57, 3.06) \).

Step 3: Calculate the work done using the dot product

The work done by the force is the dot product of the force vector and the displacement vector: \( (2.57, 3.06) \cdot (5, 3) \).Multiply the corresponding components of the vectors together and add the results to get the magnitude of work done, which is: \( (2.57 * 5) + (3.06 * 3) \approx 19.28 \) foot-pounds.