Question

(a)

A copper block with a mass of 1.6 kg initially slides over a rough horizontal surface with a speed of 6.6 m/s. Friction slows the block to rest. While slowing to rest, 85.0% of the kinetic energy of the block is absorbed by the block itself as internal energy. What is the temperature increase of the block? (Enter your answer in degrees Celsius.)
°C
(b)
What happens to the remaining energy?
It becomes chemical energy.
It is absorbed by the horizontal surface on which the block slides.
It is so minute that it doesn't factor into the equation.
It vanishes from the universe.

Answer 1

Hey, Misha! I see you're working on a physics problem for college. I'd be happy to help you out!

(a) To solve for the temperature increase of the copper block, we can use the equation:

ΔE = mcΔT

Where ΔE is the change in internal energy of the block, m is the mass of the block, c is the specific heat capacity of copper, and ΔT is the change in temperature.

First, we need to find the initial kinetic energy of the block:

KE = 1/2mv^2 = 1/2(1.6 kg)(6.6 m/s)^2 = 35.1 J

Next, we need to find the internal energy absorbed by the block:

ΔE = 0.85(KE) = 0.85(35.1 J) = 29.8 J

Finally, we can solve for ΔT:

ΔT = ΔE/(mc) = (29.8 J)/(1.6 kg)(0.385 J/kg°C) ≈ 47°C

Therefore, the temperature increase of the copper block is approximately 47°C.

(b) The remaining energy is converted into thermal energy and dissipated into the surroundings as heat. It does not vanish from the universe, but rather it is dispersed into the environment.