Thermodynamics and Mechanics questions

[nasar176]

I am self studying and I can't ask anyone else so can you please help me out with these problems?

Two blocks of iron, one of mass m at 10.0C and the other of mass 2m at 25.0c, are placed in contact with each other. If no heat is exchanged with the surroundings, which of the following is the final temperature of the two blocks?

A)10
B)15
.
D) 20C ( this is the answer)

I have no idea how to solve the above problem I don't kno, any help would be really apreciated.

Q2) A 1kg block attached to a string revolves in a vertical circle of 1 meter radius near the surface of the earth. what is the minimum speed of the block which will keep the string taut all the time?

The answer is (9.8)^1/2. I tried using the centripetal force formula and the acceleration formula but i couldn't somehow come up with an answer.

Q3) A ball, initially at rest at t 0 sec, rolls with constant acceleration down an inclined plane 10m long, if the ball rolls 1 meter in the first 2 sec, how far will it have rolled at t=4 secs?

for this question i just added the 2 four times and got the answer 8 but the book says the answers is 4 meters

 

[nasar176]

ok i figured out the 1st one, just need help with the others

 

[fgb]

For Q2, think about the "worst" case (since it asks for the minimal speed - which will barely keep the string taut). When the block is at the top of the circle, both gravity and tension contribute to the centripetal force, therefore
$F_g + T = \frac{mV^2}{R}$

In the worst case scenario (the minimal speed), the string is barely taut - which means it is barely pulling the block - so you can say T = 0 for the minimal speed. Therefore
$mg = \frac{mV^2}{R}$
$V = \sqrt{Rg}$
which results in 9.8^(1/2) :)

For Q3, the "easy way" is to remember that, for constant acceleration, $\Delta S = v_ot + at^2/2$, so as Vo = 0, $S \alpha t^2$, where S stands for the displacement.
Therefore, if you double the time, the distante will be multiplied by four (since it depends on the square of the time), which gives you 4m.

Did i make any sense? :P

 

[nasar176]

Thank you sooo much!

 

[Nickie]

Hello, I am happy to help you with these problems.

For the first problem, we can use the principle of conservation of energy to solve it. The total energy of the system (two blocks) remains constant, so we can set the initial and final energy equal to each other. The initial energy is the sum of the potential energy and the thermal energy of the two blocks, which can be expressed as:

Ei = mgh1 + mC1ΔT1 + 2mgh2 + 2mC2ΔT2

Where m is the mass of the first block, 2m is the mass of the second block, g is the acceleration due to gravity, h1 and h2 are the heights of the blocks, C1 and C2 are the specific heat capacities of iron, and ΔT1 and ΔT2 are the temperature differences of the blocks.

The final energy is just the potential energy of the two blocks at the final temperature, which can be expressed as:

Ef = (m+2m)ghf

Setting Ei equal to Ef and solving for hf, we get:

hf = [(mgh1 + mC1ΔT1 + 2mgh2 + 2mC2ΔT2) / (m+2m)g]

Substituting the given values, we get:

hf = [(m*9.8*0 + m*0.449*15 + 2m*9.8*0 + 2m*0.449*25) / (m+2m)*9.8]

Simplifying, we get:

hf = [(0.449m + 2.245m) / (3m)] = 0.898m

Therefore, the final temperature of the two blocks is 0.898 times the initial temperature of the second block, which is 25°C. So the final temperature is 22.45°C.

For the second problem, we can use the centripetal force formula to solve it. The minimum speed required to keep the string taut all the time is when the centripetal force equals the weight of the block, which can be expressed as:

Fc = mv^2/r = mg

Solving for v, we get:

v = (gr)^1/2

Substituting the given values, we get:

v = (9.8*