A cart of weight 400N is pushed up a slope by the resultant force of 50N. The cart moves a distance of 10m. By going up the slope, the change in height of the is 1m. What is the work done against friction up the slope?

The problem above is creating a slight confusion.

We can find the work done up the slope as 500J. While the work done to just lift the cart of the ground 1m as 400J. The work done against friction is the difference of the two. Hence 100J.

However, I do not understand the intuition for this calculation. When we calculate the work done up the slope, is it somehow the sum of both the work done vertically and the work done horizontally? Then we remove this work done vertically as it is independent of friction?

Is the above true? If so, how do we prove the work done at an angle is the sum of the work done horizontally and vertically? It doesn't intuitive sense considered work is a scalar.

Edit: I struggle to understand why we can just take $50\times 10=500$ to be the total work done. It seems unintuitive why the calculation is so easy. How do we prove regardless of direction that total work done is always $Fs$?

  • 1
    $\begingroup$ If this is the statement of the problem, there is no way to tell the work of friction. The work of friction can be anything from zero to 100 J, and corresponingly the gain in KE can be from 100 J to zero. $\endgroup$
    – nasu
    Commented Apr 28, 2023 at 20:03
  • $\begingroup$ Is the cart moving at constant velocity at all times? $\endgroup$ Commented Apr 29, 2023 at 18:51
  • $\begingroup$ This is unclear. What kind of "cart"? One with wheels? Is it sliding up the slope? $\endgroup$
    – Bob D
    Commented May 4, 2023 at 15:00

4 Answers 4


It is not a resultant force of 50N. It is just an applied force. If it is resultant, it would have been a resultant after cancelling friction, and thus we cannot subtract to get the frictional work done. If the question said that, the question is wrong.

You are having a misconception of what happened in the solution.

The definition of work done is force times the distance it is applied, where one of the two is being projected onto the other, i.e. get the parallel part. If you want to compute the vertical part and horizontal part, yes, you can, but nobody is doing that. It makes more sense to just do 50N x 10m to get the whole work done 500J.

Of course, if you wanted to separate that into vertical and horizontal work done, then you would have to do $$ \text{horizontal work} = ( 50N \sqrt{.99} ) ( 10m \sqrt{.99} ) = 495J \\ \text{vertical work} = ( 50N / 10 ) ( 1m ) = 5J $$ the sum of whom is again 500J.

Instead, 400J is the gain in gravitational potential energy, and that is totally different. What we are saying is that $$ \text{work done by force} = \text{GPE} + \text{work done to overcome friction} $$


Firstly work or work done is not a verctor quantity. it is a scalar quantity so whatever we calculate is net work done (+ve and -ve are conventions to show whether work is done in direction of displacement or oppposite to it). Work is not a vector quantity but it depends upon the frame of reference, since potential energy is defined with respect to a particular position.

Also when you say work done by slope - it is incorrect to say that. It is not the work done by slope, rather it is the work done by external slope. Further, the aqnswer is not conceptually calculated by subtracting them, rather it is done by using the Work energy theorem

The following problem can be solved simply using work energy theroem , which i think will provide an intuitive answer as well. Also remember that this theorem is valid regardless of nay non conservative force.

Change in Kinetic energy= Net work done

Since the body started at rest before the force was applied moved a distance of 10m and stopped. Hence. change in kinetic energy=0

Net work done= work done against gravity + work done by external force + work done against friction.

let work done by friction= x

work done against gravity= -400 (-mgh=400*1)

work done by external force = (50*10)=500

Therefore using WET

-400+500+x=0 hence x= - 100

What you mean by work done by slope is not actually that. 500 J here is the work done by external force. And that work done by external force is independent of gravity. SO it is not actually the difference how you are assuming , rather it is the sum of of or total work done. But since it is has different signs, the net resultant is the difference of the two.

Hope it helps.


Now I think the question is missing an aspect of the object being at 0 kinetic energy. With that assumption, the net work done on the object is 0. So the work done by the applied force balances the work done against gravity and friction. That will then give the work done against friction as $W_F - W_{gravity} = 50*10 - 400*1 = 100J$.

In full detail you can decompose the forces along the slope and perpendicular to the slope. If $\theta$ is the slope angle, you get $$ \Delta KE = 0 = W =(F-mg sin(\theta)-F_\mu) s = F s - mg sin(\theta)s - W_{fric}$$ To find $sin(\theta)$, you can look at more details about the slope. With s= 10m, the objects height increased by 1m, so $sin(\theta)=1/10$, so you can add that and get $ mg sin(\theta) s = 400 *(1 /10) *10 = 400$ (which is also what you would get by simply doing weight 400N going up by 1m)

This is actually more general than just the above situation. The reason being that while work done by a force is scalar, it is also the dot product of the force and displacement $ W = \vec{F}. \vec{s}$. You can compute the dot product by taking components of the force and displacement vectors in any direction and the value is the same. If you write everything in horizontal and vertical components then the work is $$W = F_h s_h + F_v s_v$$ This is similar to saying vertical work and horizontal work. Often it's easier to take the component along the displacement vector, and orthogonal to it. Then one you takes the force along the displacement which is $Fcos(\theta)$ where $\theta$ is the angle between the two vectors giving $$W = F s cos(\theta)$$.

You can go into the algebra and show these two values are same. Let by having s be at angle $\theta_1$ and F be at angle $\theta_2$ from horizontal, both in anti-clockwise direction. (for simplicity consider two dimension only). Then the first equation gives: $$W = (F cos(\theta_2)) ( scos(\theta_1)) + (F sin(\theta_2)) (s sin(\theta_1)) = Fs ( cos(\theta_2) cos (\theta_1) + sin (\theta_2) sin(\theta_1))$$

Now from trigonometry we know cos(A+B) = cosA cosB - sinA sinB, which when applied above gives $ W = F s (cos(\theta_1 - \theta_2)) $; but $\theta_2 - \theta_1$ is nothing but the angle between F and s i.e. the second equation.


the work without friction is

$$W_1=\int_0^L (F-m\,g\sin(\phi))\,ds=\left[F-m\,g\sin(\phi)\right]\,L$$

where $~\phi~$ is the slop angle and F is the force that pushed the cart up the slop.

the work with friction is: $$W_2=\int_0^L (F-m\,g\sin(\phi)-F_\mu)\,ds=\left[F-m\,g\sin(\phi)-\mu\,m\,g\cos(\phi)\right]\,L$$

with $$ \sin(\phi)=\frac hL\quad,\cos^2(\phi)=1-\left(\frac hL\right)^2\quad,m\,g=w\\ W_2=F\,L+w\left(\mu\sqrt{L^2-h^2}-h\right)$$

thus $$\Delta W=W_2-W_1=w\left(\mu\sqrt{L^2-h^2}\right)\quad ,L > h $$ where $~\mu~$ is the friction coefficient.

you need to do more work to bring the cart up the slope , if you have a road friction $~W_2 > W_1~$


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