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If you're trying to raise an object a set height, call it $h$, by pushing it up an incline, and there is a defined kinetic friction $μ$ along the incline, am I correct that you do less work if the incline is steeper? I calculate the work to be: $(mgsinθ + μgcosθ)*h/sinθ$. Conceptually, the work against gravity to raise the object will be $mgh$ no matter what, and the work against friction will be greater if you have to push the object up a longer incline, so the shorter/stepper the incline, the less work one will do. In the limiting case, the least amount of work would be if there were no incline, and you just lifted the object straight up, because you'd eliminate all the work done to overcome friction. This seems counter-intuitive to me, but correct. And I suppose it "seems" like more work to just lift the object without any incline because that would require more power. Is my analysis accurate? Any thoughts greatly appreciated.

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    $\begingroup$ The steeper incline requires more force. Power may be arbitrarily low in either case by going slowly. $\endgroup$
    – BowlOfRed
    Aug 28, 2023 at 21:03
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    $\begingroup$ @BowlOfRed The steeper incline doesn't necessarily require more force. Pulling an object across an interface with a coefficient of friction greater than one requires a force greater than the object's weight. It would take more force to pull a rubber block up a rubber ramp than it would to just lift it. $\endgroup$ Aug 28, 2023 at 21:08
  • $\begingroup$ @NuclearHoagie, It is entirely possible, using the right materials, to create an incline on which, pushing some object to the top requires more force than just simply lifting the object straight up to that same height. But that is not how people build real ramps. What they tend to do in reality, is to put the load on a wheeled cart, and then push the cart up the ramp. They still end up doing more total work than if they simply lifted the load straight up, but the maximum force needed to move the load up the ramp can be made as small as you like—only limited by how long you can make the ramp. $\endgroup$ Aug 29, 2023 at 2:17

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Your reasoning is correct, you do less work with a steeper incline.

I'm not sure why you find it counterintuitive - possibly because it implies $mgh$ does not work? If so, keep in mind that gravity is a conservative force (i.e., the amount of work you need to get from point A to point B depends only on where point A and point B are). This is why $PE = mgh$ depends only on $h$. However, friction is not a conservative force. Once you consider friction, then you get the conclusion in the OP. If the incline were frictionless, then you return to "you do the same amount of work regardless of how steep the incline is".

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  • $\begingroup$ Thank you so much. It seemed counterintuitive for simplistic, picture-thinking reasons, I suppose; at some subliminal level I was probably thinking of someone straining to lift something heavy straight up as opposed to pushing it up a gentle incline. And perhaps there's something to that when it comes to the required power in different scenarios, but I'm glad to have a clearer grasp on the work issue -- again, many thanks for the clarifying response. $\endgroup$
    – Chris
    Aug 29, 2023 at 15:07
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am I correct that you do less work if the incline is steeper?

Yes.

If the incline were frictionless, the work $W$ required to raise the object a height $h$ would simply be $W=mgh$, regardless of the angle of incline (distance traveled along the incline) since gravity is a conservative force. In the presence of a kinetic friction force $f_{f}$ , which is a non-conservative force, additional work, $W_{f}=f_{f}d$ where $d$ is the length of the incline, needs to be done. Consequently, $W_{total}= W+W_{f}$.

The steeper the incline, less additional work $W_{f}$ is needed for two reasons. (1) The friction force $f_{f}$ is proportional to the normal reaction force of the incline and the normal force decreases with increasing angle of incline thus reducing $f_{f}$. (2) The distance $d$ over which the friction force $f_{f}$ acts decreases with increasing angle of incline for a given height.

Hope this helps.

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  • $\begingroup$ It helps very much -- thank you $\endgroup$
    – Chris
    Aug 29, 2023 at 15:02

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