If there is a block on a ramp, and it slides to the bottom (with friction), then is the work done by gravity always equal to the work done against friction?

  • $\begingroup$ Do you mean "work done by friction"? Forces do work. What does "work done against friction" mean? $\endgroup$ Commented Jul 15, 2020 at 15:00
  • $\begingroup$ Since the displacement is opposite the force, wouldn't it be considered work done against friction? $\endgroup$
    – planckton
    Commented Jul 15, 2020 at 15:03
  • $\begingroup$ I think the usual way to describe it is "the work done by friction is negative". The phrase "work done against friction" makes it sound like you are focusing on other things doing the work, but here you really do mean "the work done by friction". It is much less confusing to say "work done by the force" in all cases. $\endgroup$ Commented Jul 15, 2020 at 15:05
  • $\begingroup$ @BioPhysicist, It is usual to talk of work done against friction. It is already clear that the work is done by gravity, and we are simply dividing it into parts, as described by Dale. The idea of work being done by friction, whether positive or negative, is bizarre, because friction does not cause movement (only opposes it), so friction cannot do work. $\endgroup$ Commented Jul 15, 2020 at 21:52
  • $\begingroup$ @CharlesFrancis The work done by a force $\mathbf F$ is $\int\mathbf F\cdot\text d\mathbf x$, so both gravity and friction do work here. Unless you are saying $\int\mathbf F\cdot\text d\mathbf x=0$ here for friction? What about an object sliding on a flat surface and coming to rest due to friction. You would say there is no force doing work here? What is doing the "work against friction" in that scenario? Just because work is negative doesn't mean it is nonexistent or there is work "against" it. $\endgroup$ Commented Jul 15, 2020 at 22:17

1 Answer 1


No. The work done by gravity is equal to the work done against friction plus the change in the kinetic energy.

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    $\begingroup$ Technically $W_\text{grav}+W_\text{fric}=\Delta K$, not $W_\text{grav}=W_\text{fric}+\Delta K$ $\endgroup$ Commented Jul 15, 2020 at 14:56
  • $\begingroup$ Oops, yes. A “by” vs “against” error. Now fixed, thanks! $\endgroup$
    – Dale
    Commented Jul 15, 2020 at 15:00
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    $\begingroup$ @Julia $W_\text{grav}=-\Delta U_\text{grav}$, which is true for all conservative forces $\endgroup$ Commented Jul 15, 2020 at 15:10
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    $\begingroup$ Ok thanks for the clarification! $\endgroup$
    – planckton
    Commented Jul 15, 2020 at 15:11
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    $\begingroup$ @Julia Don't forget to up vote all useful answers and to select one answer as the accepted answer if it sufficiently answers your question. $\endgroup$ Commented Jul 15, 2020 at 15:19

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