# Video: Finding the Work Done by a Force on a Body Using a Position-Time Expression

A body of mass 5 kg is moving under the action of the force 𝐅 measured in newtons. Its position vector after 𝑡 seconds is given by 𝐫 = (9𝑡³𝐢 + 8𝑡²𝐣) m. Find the work done by the force 𝐅 over the interval 0 ≤ 𝑡 ≤ 1.

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### Video Transcript

A body of mass five kilograms is moving under the action of the force 𝐅 measured in newtons. Its position vector after 𝑡 seconds is given by 𝐫 equals nine 𝑡 cubed 𝐢 plus eight 𝑡 squared 𝐣 meters. Find the work done by the force 𝐅 over the interval 𝑡 is greater than or equal to zero and less than or equal to one.

When calculating the work done by variable forces, we need to use integration. The work done in the first 𝑡 seconds is equal to the definite integral between zero and 𝑡 of the dot product of 𝐅 and 𝐯 with respect to 𝑡. The problem we have here is we know what 𝑡 is going to be. It’s going to be equal to one. But we don’t currently have a vector for velocity or force.

We do, however, have a vector for displacement or position of the object at 𝑡 seconds. And by recalling the fact that velocity is rate of change of the displacement or position of an object over a given time, we know that we can find 𝐯 by differentiating our vector for 𝐫 with respect to 𝑡.

Now we can do this for both the 𝐢- and 𝐣-components separately. So we’re going to differentiate nine 𝑡 cubed with respect to 𝑡. To achieve this, we multiply the entire term by the exponent and reduce that exponent by one. So we get 27𝑡 squared. Similarly, when we differentiate eight 𝑡 squared with respect to 𝑡, we get two times eight 𝑡, which is 16𝑡. And that’s great because we now have a vector that describes the velocity of our body. But what about the force?

Well, here we recall the equation 𝐅 equals 𝑚𝐚. Force is equal to the mass times the acceleration. It’s important to realize that this equation holds when 𝐅 and 𝐚 are vectors too. Now acceleration is the change in velocity with respect to time. So we can say that 𝐚 is the derivative of 𝐯 with respect to 𝑡.

And once again, we can differentiate the 𝐢-component and the 𝐣-component separately. The derivative of 27𝑡 squared is two times 27𝑡. So that’s 54𝑡. And then the derivative of 16𝑡 is 16. So we now have a vector that describes the acceleration of the body.

We know the body has a mass of five kilograms. So the force is equal to five times the vector for acceleration. That’s five times 54𝑡𝐢 plus 16𝐣, which is 270𝑡𝐢 plus 80𝐣 newtons.

Remember, we said that, to find the work done, we’re going to integrate the dot product of 𝐅 and 𝐯. To find the dot product of 𝐅 and 𝐯, we multiply the 𝐢-components. So that’s 270𝑡 times 27𝑡 squared. And we add the product of the 𝐣-components. So that’s 80 times 16𝑡. And we find that 𝐅 dot 𝐯 is 7290𝑡 cubed plus 1280𝑡.

The work done is then the definite integral between zero and one of 7290𝑡 cubed plus 1280𝑡 with respect to 𝑡. To integrate each term, we add one to the exponent and then divide by that new value. So the integral of 7290𝑡 cubed is 7290𝑡 to the fourth power over four. And the integral of 1280𝑡 is 1280𝑡 squared over two. This simplifies to 3645 over two 𝑡 to the fourth power plus 640𝑡 squared. Substituting 𝑡 equals one and 𝑡 equals zero, and we get 3645 over two plus 640 all minus zero, which is 2462.5.

Our force was measured in newtons. And our displacement or position was measured in meters. So the work done must be measured in joules. And the answer is 2462.5 joules.