Question

The temperature of a plastic cube is monitored while the cube is pushed 3.4 m across a floor at constant speed by a horizontal force of 20 N. The monitoring reveals that the thermal energy of the cube increases by 17 J. What is the increase in the thermal energy of the floor along which the cube slides

Answer 1

Answer:

$\Delta E_{floor} = 51 J$

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:        

$W = \Delta E_{T} = \Delta E_{cube} + \Delta E_{floor}$ (1)

Also, the work done on the cube by the horizontal force is giving by:

$W = F \cdot d$ (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:  

$\Delta E_{cube} + \Delta E_{floor} = F \cdot d$

$\Delta E_{floor} = F \cdot d - \Delta E_{cube}$

$\Delta E_{floor} = 20 N \cdot 3.4 m - 17 J$  

$\Delta E_{floor} = 51 J$

 

So, the increase in the thermal energy of the floor is 51 J.  

Have a nice day!