Video: Finding the Work Done by a Constant Force over a Given Distance

Consider a particle on which several forces act, one of which is known to be constant in time: 𝐅₁ = (3.0𝐒 + 4.0𝐣) N. As a result, the particle moves along the π‘₯-axis from π‘₯ = 0.0 m to π‘₯ = 5.0 m in some time interval. What is the work done by 𝐅₁?

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Video Transcript

Consider a particle on which several forces act, one of which is known to be constant in time: 𝐅 sub one equals 3.0𝐒 plus 4.0𝐣 newtons. As a result, the particle moves along the π‘₯-axis from π‘₯ equals 0.0 meters to π‘₯ equals 5.0 meters in some time interval. What is the work done by 𝐅 sub one?

Let’s call the work done by 𝐅 one capital π‘Š, and we can begin our solution by drawing a diagram of the particle’s movement. Now the force that we’re given, 𝐅 one, has both π‘₯- and 𝑦- or 𝐒- and 𝐣-components, so we draw an axis with two dimensions, both π‘₯ and 𝑦.

But as far as the particle’s actual motion, we’re told it moves along the π‘₯-axis from π‘₯ equals 0.0 meters to π‘₯ equals 5.0 meters. This is the displacement 𝐝 of the particle, and it’s a vector with both magnitude and direction.

At this point, let’s recall the definition for work done on an object. The work done, which is a scalar quantity, is equal to the force, a vector, times the displacement, another vector, of a particle or object.

In our case, work π‘Š is equal to 𝐅 sub one dotted with the displacement 𝐝. And from our diagram, we can see that the displacement 𝐝 is 5.0 meters in the 𝐒- or π‘₯-direction, so let’s input our values for 𝐅 sub one and 𝐝 into our equation for work.

The work done by 𝐅 sub one is equal to 3.0𝐒 plus 4.0𝐣 newtons dotted with 5.0𝐒 meters. Just like with other vectors, when we combine these two vectors through the dot product, we need to be mindful of their components. In this case, only like components multiply with one another. The 𝐒-component of 𝐅 one, it multiplies with the 𝐒-component of the displacement.

However, there is no 𝐣-component for the displacement, so the 𝐣-component of 𝐅 one does not act on the displacement. We can write that this way. We can see that the second of these terms will be zero because we’re multiplying by zero, so that only the 𝐒-component of 𝐅 one interacts with the displacement to yield work.

When we multiply 3.0𝐒 newtons by 5.0𝐒 meters, we get the result of 15 newton-meters. This is the scalar quantity to two significant figures of the work done by the vector 𝐅 one.

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