Answer:
[tex] \Delta E_{floor} = 51 J [/tex]
Explanation:
The work (W) done on the cube to be pushed across the floor is equal to the total thermal energy (ΔE) of the system:
[tex] W = \Delta E_{T} = \Delta E_{cube} + \Delta E_{floor} [/tex] (1)
Also, the work done on the cube by the horizontal force is giving by:
[tex] W = F \cdot d [/tex] (2)
where F: force applied to the cube , d: displacement of the cube
By equaling the equations (1) and (2), we can find the thermal energy of the floor:
[tex] \Delta E_{cube} + \Delta E_{floor} = F \cdot d [/tex]
[tex] \Delta E_{floor} = F \cdot d - \Delta E_{cube} [/tex]
[tex] \Delta E_{floor} = 20 N \cdot 3.4 m - 17 J [/tex]
[tex] \Delta E_{floor} = 51 J [/tex]
So, the increase in the thermal energy of the floor is 51 J.
Have a nice day!
