Object going up an inclined plane

The block has an initial velocity right before it reaches the plane. The question I was trying to solve asked what height the block would reach before it would start sliding back.

The force $f=0$, I just couldn't find a better picture. There's also a friction force pointing in the same direction as $mg\sin(\theta)$ which is also not included in the picture. You've been given the initial velocity, the friction coefficient, the angle of the incline and the mass of the block.

I know that I'm supposed to use the law of conservation of energy. Where the kinetic energy of the block before it reaches the ramp is equal to the potential energy gained at the turning point plus the energy lost due to friction.

$$E_k = E_p + W_f$$ Where $W_f$ is the work done by the friction to slow down the block to a halt. From this you can find $h$ and answer the question.

Here's where I am confused. Shouldn't the gravitational component force $mg\sin(\theta)$ be considered when calculating the work done on the block along the ramp? It has the same direction as the friction force. So shouldn't the equation actually be $$E_k = E_p + s(F_f + mg\sin(\theta))$$

Where $s$ is the hypotenuse of the triangle defined from the corner at $\theta$ to the turning point.

  • $\begingroup$ the friction force acts up the plane in the direction of $f$, not in the direction of $mg\sin\theta$ $\endgroup$
    – Bob D
    Feb 1, 2022 at 17:21
  • $\begingroup$ @BobD How so? The block is moving up the plane, friction force is always in the opposite direction in situations like this. If what you're saying is true, then the block would accelerate up the plane and never reach a stop. Am I misunderstanding your comment? $\endgroup$ Feb 1, 2022 at 23:42
  • $\begingroup$ I have edited your title so that it better reflects your question, which is more conceptual and less about this specific problem. Feel free to roll back the changes if you liked it the way it was. $\endgroup$ Feb 2, 2022 at 0:43

1 Answer 1


The work done by gravity has already been accounted for, in the potential energy term. Specifically, the change in potential energy due to gravity is defined to be minus the work done by gravity between the initial and final points. In general, if we split our forces up into conservative and non-conservative forces, we have from the work-energy theorem $$ \Delta E_k = W_\text{cons} + W_\text{non-cons} $$ and since $\Delta E_p$ is defined to be $-W_\text{cons}$, this is equivalent to $$ \Delta E_k + \Delta E_p = W_\text{non-cons}. $$

  • $\begingroup$ If $\Delta E_p= -W_\text{non-cons}$ shouldn't the last equation then be $\Delta E_k + \Delta E_p = W_\text{cons}$ ? $\endgroup$ Feb 2, 2022 at 0:18
  • $\begingroup$ @ShootinLemons: My mistake! The last equation is correct; I had a typo earlier. The key point is that potential energy is always associated with conservative forces. $\endgroup$ Feb 2, 2022 at 0:31
  • $\begingroup$ +1, but is that how potential energy is defined? I thought it was defined through a function $U$ existing such that $F=-\nabla U$, and then because of this $W=-\Delta U$. Otherwise we could say any force has a potential energy function by just slapping a negative sign on the work done by that force. $\endgroup$ Feb 2, 2022 at 14:04
  • $\begingroup$ @BioPhysicist: The criterion that there exists such a $U$ (or equivalently that $\nabla \times \vec{F} = 0$) is the definition of a conservative force. You're of course correct that we can't define a potential energy for any old force. $\endgroup$ Feb 2, 2022 at 14:18
  • 1
    $\begingroup$ @ShootinLemons The point is that you can either 1) explicitly calculate the work done by gravity, or 2) consider changes in potential energy. These are completely equivalent, but this equivalence is sometimes masked by the general equations written for these two methods, which do end up looking different before you plug in expressions for $W$ and/or $U$ $\endgroup$ Feb 2, 2022 at 16:25

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