Work done by friction but no net change in energy in this case

Question

Q) A rough inclined plane is placed on a cart moving with a constant velocity $u$ on horizontal ground. A block of mass $M$ rests on the incline. Is any work done by force of friction between the block and incline? Is there then a dissipation of energy?

In this question, what I figured out is that:

• $\theta$ is taken as angle of inclination.
• x is taken as the displacement done due to to the movement of cart and hence block w.r.t ground.
• work done by friction is $$W_{\text{friction}} = mg \sinθ \cosθ x$$ (since $f = mg \sin \theta$, no slipping case)
• work done by normal reaction is $$W_{\text{normal}} = -mg \sinθ \cosθ x$$

so the total change in energy is $0$ (zero), since work done by the non-conservative force is equal to change in mechanical energy.

So, in conclusion, is there work done by friction? So, is there then a dissipation of energy? And is my thinking correct?

Answer 1

And is my thinking correct?

Yes, your thinking is correct.

So, is there then a dissipation of energy?

There is no dissipation of energy. The same amount of mechanical energy that enters the block through static friction leaves the block through the normal force. No mechanical energy is changed to other forms of energy in the process.

So, in conclusion, is there work done by friction?

Yes, there is positive work done by friction in the amount you calculated. There is also negative work done by the normal force in the amount you calculated. Trust the math when done correctly, as you did.

This is an uncomfortable conclusion for many people, but it is correct. The static friction force can do mechanical work in any scenario where the surface is moving. For example, consider a box in an accelerating cart on a level road, the only horizontal force is the static friction force, and it accelerates the box doing work on it and increasing the KE.

Friction and the normal force always represent two components of the same interaction. In this case they are both components of the contact force with the incline. This force can be considered as a total vector or as the sum of two components, one perpendicular ($\vec F_{normal}$) and one parallel to the incline ($\vec F_{friction}$).

In this case the total force ($\vec F_{contact}=\vec F_{normal}+\vec F_{friction}$) is perpendicular to the displacement, so we know immediately that the total work will be 0. But that does not mean that the work done by an individual component will be 0, it only means that the work done by one component will be the opposite of the work done by the other component.

In other examples, like the box on the accelerating cart, the contact force is not perpendicular to the displacement so there is work done, and it is attributed specifically to the friction component of the contact force.

The bottom line is trust the math, it works out consistently when it is correctly done as you did. There is an unfortunate almost instinctive desire to say that static friction never does work, but the math does not support that. When the object is moving static friction does work, as you showed here.

Answer 2

Since the velocity of the cart is constant it is irrelevant to the problem. That means the inclined plane might as well be considered stationary.

The fact that the block is at rest on the plane simply means the maximum possible static friction force between the block and the plane hasn’t been exceeded.

No work is done by static friction between the block and the plane and no energy is dissipated.

Hope this helps.

Answer 3

There is no net work done by static friction. Let's call the bodies 1 and 2, $\mathbf{v}_1$ and $\mathbf{v}_2 = \mathbf{v}_1$ their velocity (no relative velocity since static friction acts with no sliding), and force $\mathbf{F}_{12}$ force acting on body 1 due to body 2, and $\mathbf{F}_{21} = - \mathbf{F}_{12}$ force acting on body 2 due to body 1, opposite following the 3-rd principle of Newton dynamics, the principle of action/reaction.

Evaluating the net power of the static friction here,

$$P = \mathbf{F}_{12} \cdot \mathbf{v}_1 + \mathbf{F}_{21} \cdot \mathbf{v}_2 = \underbrace{(\mathbf{F}_{12} + \mathbf{F}_{21} )}_{=0} \cdot \mathbf{v}_1 = 0 \ .$$

Remark. Here $\mathbf{F}_{ij}$ contains both normal reaction and friction as the tangential component of the force. In order to put in evidence friction, you need to write $$\mathbf{F}_{ij} = \mathbf{N}_{ij} + \mathbf{F}^{\mu}_{ij} \ ,$$

with $\mathbf{N}_{ij} = - \mathbf{N}_{ji}$ and $\mathbf{F}^{\mu}_{ij} = - \mathbf{F}^{\mu}_{ji}$ for action/reaction principle.