# Video: Finding the Work Done by a Force in Vector Form Acting on a Body Moving between Two Points

A particle moves in a plane in which π and π are perpendicular unit vectors. A force, πΉ = (9π + π) N acts on the particle. The particle moves from the origin to the point with position vector (β9π + 6π) m. Find the work done by the force.

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### Video Transcript

A particle moves in a plane in which π and π are perpendicular unit vectors. A force πΉ equals nine π plus π newtons acts on the particle. The particle moves from the origin to the point with position vector negative nine π plus six π metres. Find the work done by the force.

In this statement, weβre given a force πΉ and weβre told the components of that force. Weβre further told that the force helps to move a particle from the origin to the point negative nine π six π metres. Calling that position vector capital π, we want to solve for the work done by the force πΉ, which we can call π.

Letβs start out by drawing a diagram of the force and the position vector. If we draw in a set of coordinate axes, where π is the horizontal component and π is the vertical component, then we can graphically represent the force πΉ as well as the displacement π, which weβve drawn in moving from the origin to their respective component endpoints.

Since weβre solving for the work done π by the force πΉ, we can recall that that work is equal to the magnitude of the force πΉ times the magnitude of the displacement π multiplied by the cosine of the angle between those two vectors. In our case, we can write that π is equal to the magnitude of πΉ times the magnitude of π multiplied by the cosine of the angle between them, which we can call π.

We can start out by calculating what that angle π is. Looking at our diagram, if we call the angle between the positive π hat axis and πΉ πΌ and the angle between the negative π hat axis and π π½, then we can write that π is equal to 180 degrees minus πΌ minus π½ or converting from degrees to radians, π equals π minus πΌ minus π½.

Considering the angles πΌ and π½, we donβ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 πΉ ends at the point nine, one and the displacement vector π ends at the point negative nine, six. We can use these π, π coordinates of πΉ and π to help solve for πΌ and π½.

First, considering the angle πΌ, we can write that the tangent of πΌ is equal to one divided by nine. Thatβs the vertical component of that vector divided by its horizontal component. This indicates that πΌ is the inverse tangent of one-ninth.

Now, letβs consider the angle π½. We can write the tangent of π½ as positive six over nine β both positive because we can consider this as effectively being a first quadrant angle. So π½ equals the inverse tangent of six over nine. And we now have an expression for π that we can enter on our calculator and solve for. With that step done, letβs go back to our equation for work and consider the magnitude of the force and the magnitude of our displacement π.

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 πΉ are nine π and one π. So we take the square root of nine squared plus one squared. And that result will be equal to the magnitude of πΉ in units of newtons. And the magnitude of the displacement π will be equal to the square root of negative nine squared plus six squared, the squares of the components of π metres.

Bringing together our expressions for the magnitude of πΉ, the magnitude of π, and π, plugging these values in, and then entering them on our calculator, we find for a value of π negative 75 joules. Thatβs the work done on the particle by the force πΉ over its displacement π.