If the block has a temperature of 22.4∘ C at the top of the ramp, what is the temperature of the block when it reaches the bottom of the ramp?

Question

Suppose a 26.4 kg block of aluminum (c = 900 J/(kg⋅∘ C)) is being pushed down a steep ramp that is inclined at an angle of 59∘ from the horizontal by a 145 N force directed vertically straight down on the block. The surface of the ramp is so rough that the block moves with a constant velocity of 4.9 m/s for a distance of 146 m down the length of the ramp. You may assume that no heat is lost to the environment.

Answer 1

If you create a sketch you might come up with something like this:

Using the x-y coordinate plane on an angle (as shown) makes things a little easier.

Break the applied force (F_A) up into two components F_A-x and F_A-y like this:

F_A-x = 145sin(59°) = 124.29 N

F_A-y = 145cos(59°) = 74.68 N

The weight of the block is (26.4 kg)(9.81 m/s²) = 258.984 N and this weight can also be broken up into the same components Wx and Wy:

W_x = 258.984sin(59°) = 221.99 N

W_y = 258.984cos(59°) = 133.39 N

So the total force in the x-direction is F_A-x + W_x = 124.29 + 221.99 = 346.28 N

Since the block is traveling at a constant speed, the acceleration is zero and since F = ma, the net force must be zero. That means there must be a frictional force, equal and opposite of 346.28 N.

The work done on the block is the frictional force multiplied by the distance traveled.:

W = Fd = (346.28)(146) = 50557 joules.

We are told that all that energy goes into heating the block.

Q = mCpΔT

50557 = (26.4)(900)(ΔT)

ΔT = 2.13 °C

So the final temperature of the block is 24.5 °C