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So recently a high-schooler asked me this question.

Consider a particle (of mass "m") moving with 10m/s (in the ground frame) in a certain direction on a frictionless surface. Now along its path it encounters a finite region of non-zero frictional force and once it comes out of that region , say its velocity is 6m/s(in the ground frame again) in the same direction.

So from the Work-Energy principle we know that the work done by friction will be $$W_f=\frac{m}{2}*(10^2-6^2)=32m$$ Now let us have an inertial frame moving with 2m/sec(wrt. ground) in the opposite direction. From this frame, the particle's initial speed is 12m/s and final speed is 8m/s. So again from work-energy principle we have that $$W_f=\frac{m}{2}*(12^2-8^2)=40m$$ So an excess energy of 8m is lost somehow ! So did friction have done extra work here ??

I am sure work done by friction is indpendent of the frame of reference and that's what I told him too but to be honest I don't really know where exactly is this extra energy going to ?

Edit : Just to clarify why my question is different from the one linked is that I accept that work done could be frame dependent (which is what the answers in the duplicate one are trying to say). I didn't clearly mention it earlier but my main issue was that the work done by friction is lost as heat so doesn't that imply that there would be more heating in the moving frame (more temperature change) ?? I expected that someone will talk about this in their answer.

Dale
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Ankit
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1 Answers1

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The extra energy goes into the kinetic energy of the earth.

The mix of a variable $m$ and a numerical $10 \mathrm{\ m/s}$ and $6 \mathrm{\ m/s}$ is inconvenient, so I will use only variables.

This is easier to analyze with power rather than with work. If work is absolutely needed then you will need to integrate the power over time. The mechanical power of a force is $$P=\vec f \cdot \vec v$$ where $\vec v$ is the velocity of the material at the point where the force $\vec f$ is applied. A positive $P$ means mechanical energy is going in to the system, and a negative force means mechanical energy is leaving the system.

Take the positive $x$ direction as the direction of travel of the block in the ground frame. So while the block is on the rough patch the velocity of the block $\vec v_B=v_B \ \hat i$ and the frictional force on the ground is $\vec f_B = -f_B \ \hat i$. So the mechanical power of the frictional force on the block is $$P_B=\vec f_B \cdot \vec v_B =-f_B \ v_B$$ This is a negative number, meaning that mechanical power is leaving the block.

Now, consider the ground. In the ground's frame the ground's velocity is $\vec v_G=0 \ \hat i$ and the force acting on the ground is $\vec f_G=-\vec f_B=f_B \ \hat i$. So the mechanical power of the frictional force on the ground is $$P_G=\vec f_G \cdot \vec v_G = 0$$ meaning that none of the mechanical power leaving the block enters the ground.

So there is mechanical power leaving the block, but not entering the ground. This power represents mechanical energy that is lost to heat. The heat generated at the interface between the block and the ground is $\dot Q=-(P_B+P_G)=f_B \ v_B$.

Now, in an inertial frame moving at velocity $\vec u$ the forces are $\vec f'=\vec f$ and the velocities are $\vec v'=\vec v - \vec u$. So the mechanical powers are $$P'=\vec f' \cdot \vec v'=\vec f \cdot (\vec v -\vec u)=P-\vec f \cdot \vec u$$

So we can simply calculate $$P'_B=-f_B \ v_B - \vec f_B \cdot \vec u=P_B-\Delta P$$ and $$P'_G=0 - \vec f_G \cdot \vec u = \vec f_B \cdot \vec u=P_G+\Delta P$$ where $\Delta P=\vec f_B \cdot \vec u$.

Depending on the direction of $\vec u$ we can have $\Delta P$ be positive, negative, or zero. This means that we can have more or less mechanical energy leaving the block due to friction in the moving frame.

The heat generated at the interface between the block and the ground is $$\dot Q’=-(P'_B+P'_G)= P_B -\Delta P + P_G + \Delta P=f_B \ v_B=\dot Q$$ So the heat produced at the interface is the same in both frames. The difference in power between the two scenarios goes into mechanical work done on the ground. In other words, in the ground frame the KE of the ground is unchanged, but in the moving frame the KE of the ground changes. The mechanical power is different in the two frames, but the mechanical power is not directly the heat. The heat is equal to the total loss of mechanical energy. That quantity is the same in both frames $\dot Q=\dot Q’$

Dale
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