Why Is Work the Same for Different Distances on an Inclined Plane?

AI Thread Summary
Work is calculated using the formula W = fd, and in this discussion, it is noted that work remains consistent regardless of the distance traveled on an inclined plane, whether d = 4 or d = 0.5. The reasoning is that increasing the distance reduces the necessary force by a factor of 8, leading to the conclusion that work remains the same. The question arises about the horizontal component of force when pushing an object up the ramp, suggesting a potential misunderstanding of the triangle labeling in the problem. The minimum force required to lift an object vertically is compared to the force needed to move it up the ramp, confirming that the forces relate as F2 = F1/8. This explanation clarifies the relationship between force, distance, and work on an inclined plane.
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W = fd
The answer is 8. The solution manual says work is the same for either scenario, d = 4 or d = 0.5

From this assumption I can see that by increasing distance from 0.5 to 4, you reduce the force necessary by a factor of 8.

However what I cannot comprehend is: where does the assumption that work is the same for both scenarios come from? Since you have the added horizontal component to consider when pushing it up the ramp, isn't that extra work? Or...Have I labeled the triangle incorrectly? Should the 4 be labeled on the other leg rather than the hypotenuse?Sorry if this is a silly question but I haven't taken intro physics in a couple years and am currently studying for the mcat. It seems not all of the principles of physics have hstayed with me lol. Thank u in advance pf!

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The Attempt at a Solution

 
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You can analyze this problem in this way

The min force required to lift an object of mass m vertically up from bottom to top of the step is F1= mg

The min force required to move an object of mass m up along the ramp from bottom to top of the step is F2= mgsinθ

From this you can deduce F2=F1/8 .

Does this help ?
 
Yasssss! Thank you so much!
 
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