Question Video: Finding the Work Done by a Force Given in Vector Form Acting on a Body given Its Displacement–Time Expression Mathematics

A particle moves in a plane in which 𝐒 and 𝐣 are perpendicular unit vectors. Its displacement from the origin at time 𝑑 seconds is given by 𝐫 = [(2𝑑² + 7)𝐒 + (𝑑 +7)𝐣] m. And it is acted on by a force 𝐅 = (6𝐒 + 3𝐣) N. How much work does the force do between 𝑑 = 2 s and 𝑑 = 3 s?

02:25

Video Transcript

A particle moves in a plane in which 𝐒 and 𝐣 are perpendicular unit vectors. Its displacement from the origin at time 𝑑 seconds is given by 𝐫 equals two 𝑑 squared plus seven 𝐒 plus 𝑑 plus seven 𝐣 meters. And it is acted on by a force 𝐅 equals six 𝐒 plus three 𝐣 newtons. How much work does the force do between 𝑑 equals two seconds and 𝑑 equals three seconds?

Since work done is the dot product of the vector force and the vector for displacement, we’re going to need to calculate the displacement of this particle. We’re given its displacement at some time 𝑑. But we’re also asked how much work the force does between 𝑑 equals two seconds and 𝑑 equals three seconds. So, let’s substitute 𝑑 equals two and 𝑑 equals three into our expressions for the displacement. Beginning with 𝑑 equals three seconds, we get two times three squared plus seven 𝐒 plus three plus seven 𝐣. That’s 25𝐒 plus 10𝐣. And we’re working in meters. So the displacement from the origin and, hence, the position of the particle at three seconds is 25𝐒 plus 10𝐣 meters. In a similar way, we can find the position of the particle at two seconds by substituting 𝑑 equals two into our expression. And we get 15𝐒 plus nine 𝐣 meters.

The displacement then of the particle between these two times is the difference here. It’s 𝐝 sub three minus 𝐝 sub two. That’s 25𝐒 plus 10𝐣 minus 15𝐒 plus nine 𝐣. And of course, we simply subtract the individual components. 25 minus 15 is 10, and 10 minus nine is one. So, the displacement is simply 10𝐒 plus 𝐣, and that’s in meters. Since the work done is the dot product of the force and displacement, we need to find the dot product of six 𝐒 plus three 𝐣 and the displacement we just calculated. It’s the dot or scalar product of six 𝐒 plus three 𝐣 and 10𝐒 plus 𝐣. That’s six times 10 plus three times one, which is, of course, equal to 63. The work done then is equal to 63 joules.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.