Question concerning work and horizontal force.

AI Thread Summary
A horizontal force of 300N is required to push a stalled automobile at a constant speed over a distance of 5.0m, resulting in a work calculation of 1.5 x 10^3 J using the formula W = F x Δx. The discussion clarifies that while both W = F x Δx and W = F s cos θ yield the same result, the latter is a more general formula applicable in various scenarios. It emphasizes that W = F s cos θ is valid even when the force and motion are not aligned, with θ being 0 when they are. The explanation provided helps clarify the use of these formulas, enhancing understanding beyond the textbook's explanation. Understanding the context of these equations is essential for applying them correctly in physics problems.
Jim01
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Homework Statement



If it takes a horizontal force of 300N to push a stalled automobile along a level road at a constant speed, how much qwork must you do to push this automobile a distance of 5.0m?



Homework Equations



W = F x \Deltax


The Attempt at a Solution



I used the above formula W = (300N)(5.0m) = 1.5 x 103 J, however, in checking my work with the solutions manual, they used the formula W=FsCOS\theta. Both formulas come upi with the same answer.

I do not understand why the authors chose to use the W=FsCOS\theta formula because I thought that this formula was only used when "the motion of the particle and the force are not along the same line" (from the book), such as "along some arbitrary curved path."

Is this a case where W = F x \Deltax can only be used in cases where the motion of the particle and the force are along the same line but W=FsCOS\theta can be used in either case?
 
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Yes, you are correct. The most general formula for work done on a body by a force is "the product of the force and the displacement of the body in the direction of the force"; in other words, the dot product of force and displacement, which can be represented as W = F\,s\, cos \theta (strictly speaking, |F||s|cos\theta)

In the case where the direction of the force is in the same direction as the displacement, we realize that \theta is in fact 0. Hence, we obtain
W = F\,s\,cos 0 = F\,s
 
Fightfish said:
Yes, you are correct. The most general formula for work done on a body by a force is "the product of the force and the displacement of the body in the direction of the force"; in other words, the dot product of force and displacement, which can be represented as W = F\,s\, cos \theta (strictly speaking, |F||s|cos\theta)

In the case where the direction of the force is in the same direction as the displacement, we realize that \theta is in fact 0. Hence, we obtain
W = F\,s\,cos 0 = F\,s


Thank you for the explanation. You explained it much better than my textbook did.
 
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