A 0.1-kg aluminum plate,initially at 20°C,slides down a 14-m-long surface,inclined at a 29° 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⋅°C. )

Question

A 0.1-kg aluminum plate,initially at 20°C,slides down a 14-m-long surface,inclined at a 29° 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⋅°C. )
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Quizplus · Accepted Answer

First, let's find the gravitational force component down the plane:
$F_g = mg\sin	heta = 0.1 kg \cdot 9.81 m/s^2 \cdot \sin29° \approx 0.05 N$
Since the force of kinetic friction is equal and opposite, we can find the force acting on the plate down the plane:
$F = 0.05 N$
The work done by this force over the 14-m length is:
$W = Fd = 0.05 N \cdot 14 m = 0.7 J$
90% of the mechanical energy is absorbed by the aluminum, so the aluminum gains:
$\Delta E = 0.9 W = 0.63 J$
Using the specific heat capacity of aluminum, we can find the corresponding temperature increase:
$\Delta T = \frac{\Delta E}{mc} = \frac{0.63 J}{0.1 kg \cdot 900 J/kg⋅°C} \approx 0.07 °C$
Therefore, the answer is B.

A 01-Kg Aluminum Plate,initially at 20°C,slides Down a 14-M-Long Surface,inclined at at a 29°