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 ofgravity down the plane so that the plate, once started, glides down at constant velocity. If90% of the mechanical energy of the system is absorbed by the aluminum, what is itstemperature increase at the bottom of the incline? (Specific heat for aluminum is 900J/kg⋅°C.) Why do I multiply 15 by sin30?

Answer 1

A scheme of a the given situation is shown below:

First, consider that the work over the plate is done only by the component of the weight parallel to the incline (due to the perpendicular component is balanced by the friction force), then, the work on the plate is:

W = m*g*d*sinθ

where,

m: mass = 0.2kg

d: length of the incline = 15m

g: gravitational acceleration constant = 9.8m/s^2

θ = 30

By replacing the previous values into the expression for W, you obtain:

W = (0.2 kg)(9.8 m/s^2)(15 m)sin(30)

W = 14.7 J

Now, take into account that the amount of heat absorbed by the aluminum plate is given by the following formula:

Q = m*c*ΔT

Q: heat

m: mass

c: specific heat

ΔT: change in tempetaure

Take into account that the 90% of the mechanical energy is absorbed by the plate, which means that 0.9 of the work is converted to absorbed heat by the plate.

Then, you can write:

0.9W = Q

0.9(14.7J) = Q

13.23J = Q

Replace the given expression for Q into the previous equation and solve for ΔT, as follow:

m*c*ΔT = 13.23 J

ΔT = 13.23J/(m*c)

Now, replace the values of m and c for aluminum and simplify:

ΔT = 13.23J/(0.2kg*900J/kg°C)

ΔT = 0.0735°C

Hence, the temperature increase at the bottom of the incline is approximately 0.07°C