Question

The temperature of a plastic cube is monitored while the cube is pushed 3.0 macross a floor at constant speed by a horizontal force of 15 N. The thermal energy of the cube increases by 20 J. What is the increase in the thermal energy of the floor along which the cube slides?

Short answer

Increase in thermal energy of floor,$E_{th\left(floor\right)}=25\mathrm{J}$

Step-by-step solution

Given

1. Displacement of the cube d = 3.0 m
2. Force applied to the cube$F_{app}=15N$
3. increase in thermal energy of cube$∆E_{th\left(cube\right)}=20\mathrm{J}$

To understand the concept

Here the block and floor form a system. Use the concept that the total increase in thermal energy of the system is the sum of the increase in thermal energy of the cube and the floor.

Formula:

$∆E_{th\left(system\right)}=∆E_{th\left(cube\right)}+∆E_{th\left(floor\right)}$

Calculate the increase in the thermal energy of the floor along which the cube slides

Work done by applied force:

$$\begin{gathered} W_{app}=Fd\cos \varphi \\ {W}_{app}=15\times 3\cos 0 \\ {W}_{app}=45\mathrm{J}\left(\mathrm{i}\right) \\ {W}_{app}=increaseintheheatenergyofthesystem \\ \mathrm{Using}\mathrm{equation}\left(\mathrm{i}\right)\mathrm{and}\mathrm{given}\mathrm{value}, \\ {W}_{app}=∆E_{th\left(cube\right)}+∆E_{th\left(floor\right)} \\ 45=20+∆E_{th\left(floor\right)} \\ ∆E_{th\left(floor\right)}=45-20 \\ ∆E_{th\left(floor\right)}=25\mathrm{J} \end{gathered}$$