Suppose the temperature of a plastic cube is monitored while the cube is pushed \[3.0m\] across a floor at constant speed by a horizontal force of \[15N\]. The thermal energy of the cube increases by \[20J\]. What is the increase in the thermal energy of the floor along the cube side ?

Hint: In order to answer this question we must know about thermal energy. To solve this question first we should calculate the whole work done by the cube and then calculate the increasing amount of thermal energy of floor by using this formula : \[W=\Delta K+\Delta ETH(Cube)+\Delta ETH(floor)\]

Complete step-by-step solution:
Thermal energy is originated by a body due to the movement of molecules within the body. It is a one type of kinetic energy. from the above question we get –
Force applied on cube \[=15N\]
Displacement by the cube \[=3m\]
Now the work done by the cube \[=15\times 3=45J\]
as the cube is moving with a constant velocity then the kinetic energy \[\Delta K=0\] and increasing rate of thermal energy on cube is \[20J\]
 now applying the law of thermodynamics we get –
\[W=\Delta K+\Delta ETH(Cube)+\Delta ETH(floor)\]
\[or,45=0+29+\Delta ETH(floor)\]
\[or,\Delta ETH(floor)=25J\] where, \[\Delta ETH(Cube)\] is the thermal energy of a cube and \[\Delta ETH(floor)\] is the thermal energy on the floor.
Therefore, the increase of the thermal energy of floor along the cube side is \[25J\]

Note: Transformation of thermal energy to another energy is too difficult. Kinetic energy is measured by the mass and speed of that object.