I work at a STEM afterschool center, and we're designing an activity where students measure force and work on two inclined planes with the same height but different lengths.

The intention is for them to multiply the forces by the distance of each ramp and find that the product is the same number because work should be constant if both ramps have the same height. We want them to notice the tradeoff that if you want to use less force, you can build a less steep ramp, but you'll have to apply your force for a greater distance. So work remains constant. Etc...

However, we're finding that if we multiply the forces we measure (using educational scales off of Amazon. They have been calibrated) by the lengths of each ramp, we get completely different numbers. I set out to try figure out if we're doing something wrong by looking at the math.

Doing the free-body diagram and assuming friction, the force for the short and long ramps should look like

$$ F_s = mgsin(\theta_s) + \mu_k mgcos(\theta_s), $$

$$ F_l = mgsin(\theta_l) + \mu_k mgcos(\theta_l). $$

And then we force the work (no pun intended?) to be equal in both ramps

$$ W= F_s d_s = F_l d_l, $$

where $d_s$ and $d_l$ are the lengths of the short and long ramps.

All the terms in this system are known except for $\mu_k$ so, I set out to solve for it. Plugging everything into one equation looks like

$$ [mgsin(θs)+μ_kmgcos(θs)]\frac{h}{sin(\theta_s)} = [mgsin(θl)+μ_kmgcos(θl)]\frac{h}{sin(\theta_l)}. $$

Factoring $mgh$ and distributing the sine in the denominator, the $mgh$ terms end up canceling out (which is interesting, as $W = mgh$).

$$ 1+\frac{\mu}{tan(\theta_l)} = 1+\frac{\mu}{tan(\theta_s)} $$

and even $\mu_k$ ends up going as well.

$$ \frac{1}{tan(\theta_s)} = \frac{1}{tan(\theta_l)}. $$

We simply end up with $\theta_s = \theta_l$. So I guess that the work will be the same only if the ramps are the same?

At the end of the day, $W = mgh$, and it's independent of the steepness, and you can derive this when no friction is assumed and you have

$$ W_s = F_sd_s $$

$$ W=sin(\theta)mg\frac{h}{sin(\theta)}, $$

$$ W = mgh. $$

But why does this not work with friction introduced? This would look something like

$$ W = mgh\Big(1+\frac{\mu}{tan(\theta)}\Big). $$

Maybe work is not constant when accounting for friction? Does something like the heat generated and the energy lost by the friction make it so that $F_sd_s \neq F_ld_l$? So small imperfections in the real world make this experiment impractical for a high school classroom? If so, any other ideas as to how I could implement an intuitive and hands-on explanation of work and its relationship with force?

Would greatly appreciate your help!

Thanks in advance!

  • $\begingroup$ Are you determined to do this experiment with two different ramps? It is easily possible to do this experiment with only one ramp. If done that way, the math is much more straight forward and the experiment and its results will be somewhat easier to explain. $\endgroup$ Commented Jul 9, 2023 at 2:04

1 Answer 1


Assume that two blocks are sliding down two different ramps of the same height but different distances because each ramp is elevated at a different angle (no rotation is involved). An analysis of the work done by friction on each ramp follows.

Variables used:

$g$ is the acceleration due to gravity

$m$ is the mass of a given block

$h$ is the height of each ramp (constant)

$\mu_k$ is the coefficient of kinetic friction (assumed constant between ramps)

$F_f$ is the friction force impeding each sliding block

$W_f$ is the work done by friction

$L$ is the length of the hypotenuse, aka the length that each block slides

$\theta$ is the angle of each ramp relative to horizontal

Equation development for work done by friction:




Therefore, $W_f=\mu_kmgcos\theta\frac{h}{sin\theta}=\frac{\mu_kmgh}{tan\theta}$

As seen, the work done by friction varies with the angle of the ramp, so your starting assumption that friction does the same amount of work on each block is not correct. This is no doubt where your confusing results are coming from.


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