A 0.2-kg aluminum plate,initially at 20 $\degree$C,slides down a 15-m-long surface,inclined at a 30 $\degree$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 900 J/kg* $\degree$C. )
A)0.16 C $\degree$
B)0.07 C $\degree$
C)0.04 C $\degree$
D)0.03 C $\degree$

Question

Quizplus · Accepted Answer

The mechanical energy of the system is converted into internal energy of the aluminum plate, given by:
ΔE = 0.9(mgh) = mcΔT
where m = 0.2 kg is the mass of the aluminum plate, g = 9.8 m/s^2 is the acceleration due to gravity, h = 15 sin 30° = 7.5 m is the height of the incline, c = 900 J/kg°C is the specific heat of aluminum, and ΔT is the temperature increase of the aluminum plate.
Solving for ΔT, we get:
ΔT = 0.9(mgh)/(mc) = 0.9gh/c
ΔT = 0.9(0.2 kg)(9.8 m/s^2)(7.5 m)/(900 J/kg°C) ≈ 0.07°C
Therefore, the temperature increase of the aluminum plate at the bottom of the incline is 0.07°C, which is closest to option B.

A 02-Kg Aluminum Plate,initially at 20 °\degree° C,slides Down a 15-M-Long Surface,inclined at a 30

A 02-Kg Aluminum Plate,initially at 20 $\degree$ C,slides Down a 15-M-Long Surface,inclined at a 30