Question

.A 0.2-kg aluminum plate, initially at 20°C, slides down a 15-m-long surface, inclined at a 30 angle to the horizontal. The force of kinetic friction exactly balances the component of gravity down the plane so that the plate, once started, glides down at constant velocity. If 90% of the mechanical energy of the system is absorbed by the aluminum, what is its temperature increase at the bottom of the incline? (Specific heat for aluminum is 900J/kg⋅°C.)
a. 0.16 C°b. 0.07 C°c. 0.04 C°d. 0.03 C°

Answer 1

To solve this problem, it is necessary to apply the concepts related to Work according to the Force and distance, as well as the concepts related to energy lost-or gained-by heat. Mathematically the energy corresponding to heat is given as:

$Q = mC_p(\Delta T)$

Where,

m = mass

$C_p$= Specific heat

$\Delta T$ = Change in Temperature

At the same time the Work made by the Force and the distance is given as:

$W = F*d \rightarrow W=mg*d$

As the force is applied at an angle of 30 degrees, the efficient component would be given by the vertical then the work / energy would be determined as:

$W = mg*dsin(30)$

$W = (15m)(0.2kg)(9.81)(sin30)$

$W = 13.24J$

Now this energy is used to heat the aluminum. We can find the change at the temperature as follow:

$Q = mC_p(\Delta T)$

$13.24 = (0.2)(900)(\Delta T)$

$\Delta T = 0.0736 \°C$

Therefore the correct answer is B.