Now we do $-ve$ work on the system while stopping it so the work done by the system is $+ve$. Is this part right?
Not always; you've stumbled upon a subtle point in the definition of work in mechanics, which is rarely discussed. Generally, if body $A$ has done work $W$ on body B, it does not follow that the body $B$ has done work $-W$ on the body $A$. This is true only if the two material points under action of the mutual forces have moved with the same velocity.
Explanation:
Definition of rate of work: When body S acts with force $\mathbf F$ on body $\mathbf{B}$ that has velocity $\mathbf v_B$ (more accurately, it is the velocity of the mass point which experiences the force), the rate of work being done by S on $\mathbf{B}$ is defined to be
$$
\mathbf F\cdot\mathbf v_B.
$$
Notice that the velocity is that of the receiver, not that of the body the force is due to. So if I scratch my desk while the scratched portion remains at rest, no work has been done on the desk.
Thus, the definition of work is based on:
- the force due to the giving body;
- the velocity of the receiving body (its mass point the force acts on).
What about the work done on the body $S$? This is, per definition,
$$
-\mathbf F\cdot\mathbf v_S.
$$
The forces have the same magnitude and opposite signs (due to Newton's 3rd law), but there is no general relation between the two velocities of mass points where the forces are acting. If $\mathbf v_S$ is not equal to $\mathbf v_B$ during the whole process, it is possible the two works done on the bodies will not have the same magnitude.
⁂
Let's look at some specific cases.
Case 1. If a massive block $\mathbf B$ is brought to rest by another moving body $S$ with no sliding friction occurring (if the mass points of bodies experiencing the mutual forces always move with the same velocity), the velocities $\mathbf v_B,\mathbf v_S$ are the same and the two works have the same magnitude and opposite sign.
This is the case, for example, when the block is stopped in its motion by a spring mounted on a wall, or a person stops it gradually by hand. The kinetic energy of the block $E_k$ decreases to zero and equal amount of energy is added through work to the total energy of the stopping body. No heat transfer and no change in temperature need to occur, if there is no sliding friction and no temperature differences beforehand.
Case 2. If the block is stopped by forces of sliding friction - say, due to the ground - the description in terms of work is different. The mass point where the force due to ground acts on the block $\mathbf B$ is part of the block and is moving. Therefore the ground is doing work on the block (and from the reference frame of the ground, this work is negative). However, since the ground is not moving at all, the block does zero work on the ground!
This may look like violation of energy conservation, because block is slowing down without ground receiving energy, for there is no work received.
But it is only violation of mechanical energy conservation, which is fine and occurs daily. Total energy may still remain conserved, because it includes also internal energy of the block and internal energy of the ground, which change during the process.
As the block slows down, its kinetic energy transforms into different form of energy: internal energy of both the ground and the block. This happens with greatest intensity in the two faces that are in mutual mechanical contact.
The faces get warmer and for the rest of the system, they act as heat reservoirs.
The energy is transferred via heat both upwards into the block and downwards into the ground.
Further , isn't work done always equal to change in kinetic energy (by the work energy theorem) ?
Only if the only energy that changes is kinetic energy. Generally, the work-energy theorem includes other energies. Kinetic energy can change into potential energy (in gravity field, in a spring) or into internal energy (inside matter, may manifest as increase in temperature or other change of thermodynamic state).
Best Answer
With no heat input (Q=0) and KE and PE of the system =0 it can't do work on the surrounding.. note: rise in temperature, rise in internal energy is in fact rise in KE of gas molecules in a P-V-T system.
Don't complicate things, the 1st law is simply conservation of energy.. if you do work then you lose energy; as simple as that. The example you gave is a bit misleading, there the body temperature increases because of increased metabolism and cellular respiration, that energy comes from breaking of carbohydrates and thus the energy stored as chemical bonds now partly goes in pushing the object and is partly released as heat.. you can think of it like Total internal energy in form of bonds in glucose ==> excess heat + work. So here in fact Q is negative as body is giving out heat, W is positive, and internal energy decreases as some energy from the burger you had a couple of days ago is now gone..!! Don't confuse internal energy and temperature, they sometimes go hand in hand in simple systems but are completely different things. Look for the source of all energies, which form of energy is converted to what other form, and remember energy can neither be created nor destroyed (in a weaker sense, STR says something more serious) as that's all you need for the 1st law of thermodynamics.