A particle moves in a plane in which 𝑖 and 𝑗 are perpendicular unit vectors. A force 𝐹 equals nine 𝑖 plus 𝑗 newtons acts on the particle. The particle moves from the origin to the point with position vector negative nine 𝑖 plus six 𝑗 metres. Find the work done by the force.
In this statement, we’re given a force 𝐹 and we’re told the components of that force. We’re further told that the force helps to move a particle from the origin to the point negative nine 𝑖 six 𝑗 metres. Calling that position vector capital 𝑃, we want to solve for the work done by the force 𝐹, which we can call 𝑊.
Let’s start out by drawing a diagram of the force and the position vector. If we draw in a set of coordinate axes, where 𝑖 is the horizontal component and 𝑗 is the vertical component, then we can graphically represent the force 𝐹 as well as the displacement 𝑃, which we’ve drawn in moving from the origin to their respective component endpoints.
Since we’re solving for the work done 𝑊 by the force 𝐹, we can recall that that work is equal to the magnitude of the force 𝐹 times the magnitude of the displacement 𝑑 multiplied by the cosine of the angle between those two vectors. In our case, we can write that 𝑊 is equal to the magnitude of 𝐹 times the magnitude of 𝑃 multiplied by the cosine of the angle between them, which we can call 𝜃.
We can start out by calculating what that angle 𝜃 is. Looking at our diagram, if we call the angle between the positive 𝑖 hat axis and 𝐹 𝛼 and the angle between the negative 𝑖 hat axis and 𝑃 𝛽, then we can write that 𝜃 is equal to 180 degrees minus 𝛼 minus 𝛽 or converting from degrees to radians, 𝜃 equals 𝜋 minus 𝛼 minus 𝛽.
Considering the angles 𝛼 and 𝛽, we don’t know what they are, but we do know the endpoints of the vectors that start at the origin and help to define those two angles. The force vector 𝐹 ends at the point nine, one and the displacement vector 𝑃 ends at the point negative nine, six. We can use these 𝑖, 𝑗 coordinates of 𝐹 and 𝑃 to help solve for 𝛼 and 𝛽.
First, considering the angle 𝛼, we can write that the tangent of 𝛼 is equal to one divided by nine. That’s the vertical component of that vector divided by its horizontal component. This indicates that 𝛼 is the inverse tangent of one-ninth.
Now, let’s consider the angle 𝛽. We can write the tangent of 𝛽 as positive six over nine — both positive because we can consider this as effectively being a first quadrant angle. So 𝛽 equals the inverse tangent of six over nine. And we now have an expression for 𝜃 that we can enter on our calculator and solve for. With that step done, let’s go back to our equation for work and consider the magnitude of the force and the magnitude of our displacement 𝑃.
Recalling that the magnitude of a vector is equal to the square root of the sum of the squares of its components, the components of 𝐹 are nine 𝑖 and one 𝑗. So we take the square root of nine squared plus one squared. And that result will be equal to the magnitude of 𝐹 in units of newtons. And the magnitude of the displacement 𝑃 will be equal to the square root of negative nine squared plus six squared, the squares of the components of 𝑃 metres.
Bringing together our expressions for the magnitude of 𝐹, the magnitude of 𝑃, and 𝜃, plugging these values in, and then entering them on our calculator, we find for a value of 𝑊 negative 75 joules. That’s the work done on the particle by the force 𝐹 over its displacement 𝑃.