1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Thermodynamics and Mechanics questions

  1. Jan 22, 2013 #1
    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?

    D) 20C ( this is the answer)

    I have no idea how to solve the above problem I dont 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
  2. jcsd
  3. Jan 22, 2013 #2
    ok i figured out the 1st one, just need help with the others
  4. Jan 22, 2013 #3


    User Avatar

    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
    [itex]F_g + T = \frac{mV^2}{R} [/itex]

    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
    [itex] mg = \frac{mV^2}{R} [/itex]
    [itex] V = \sqrt{Rg} [/itex]
    which results in 9.8^(1/2) :)

    For Q3, the "easy way" is to remember that, for constant acceleration, [itex] \Delta S = v_ot + at^2/2[/itex], so as Vo = 0, [itex]S \alpha t^2[/itex], 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
    Last edited: Jan 22, 2013
  5. Jan 22, 2013 #4
    Thank you sooo much!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook