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Im Despret. Grade 12 Physics Question. PLEASE HELP

  1. Oct 3, 2009 #1
    Hello, Im new. Im soo despret. I want to become an Architect but I'm dumb. I aws doing fine getting 84% in physics university 11, but now, im taking physics 12 university and I don't get the answers right 65% of the time. Im so despret for help. My parents cannot afford to get me a tutor. At least half of my classmates gets nice marks on the test, but Im soo stupid that i failed the first test. And yes, I did do all my homework. I feel soo stupid its ridiculous. I need serious help.

    Question 1) Block A and B are connected by a string passing an essentially frictionless pulley. When the blocks are in motion, Block A experiences a force of kinetic friction of magnitude 5.7N. If Ma= 2.7kg and Mb=3.7kg, calculate the magnitude of

    a) The acceleration of the blocks
    b) The tension in the string.

    I have one more question but i think im asking too much. PLEASE HELP.
     
  2. jcsd
  3. Oct 3, 2009 #2
    hi there
    Your question gives inadequate information, whether blocks are on a table or not. But from the available data, it is reasonable to assume that block A is on a table, block B is hanging around the pulley - those are my assumptions.
    Under my assumptions, look at the block A first. There is friction(f) and tension(T) in the string. This gives: T - f = Ma*a ___________ 1
    Again, look at the block B. There is its weight, Mb*g and tension. This gives:
    Mb*g - T = Mb*a __________ 2
    Here the two tensions for A and B are equal because the pulley has no friction and acceleration of A and B are equal because they move together - there is no relative velocity between those two.
    Solving 1 and 2 gives both tension and acceleration.
    Good luck with your studies, Arliena.
    Thanks
     
  4. Oct 3, 2009 #3
    Hello, Thank you SOO mch for your reply! Im very sorry for my poor discription. And yes, you're right. The block A is on the table. However, I don't understand how i can get the acceleration by using (MB)(g)-T=(MB)(A) because i don't know both Acceleration and the Tension. The texbook says the Answers are:

    A) 4.8m/s*2
    B) 19N

    but I don't know how you got them. 3978375339_36c5754d0d.jpg
     
    Last edited: Oct 3, 2009
  5. Oct 3, 2009 #4
    hi there
    In the first eq, you know friction and mass of A. For the second eq, you know mass of B and g. This gives you a two simultaneous eq's in two unknowns which I think is easy enough to solve. You have two eq's, not one.
     
  6. Oct 3, 2009 #5
    I got them by separately using Newton's second law for each block.
    net force = mass times acceleration
     
  7. Oct 3, 2009 #6
    Hello, Thanks again for your reply! I tried for A) using (MB)(g)-T=(MB)(A)
    (3.7)(9.8)down-(5.7)=(3.7)(a) to (3.7)(9.8)/(3.7) - 5.7 Since 3.7 cancel, Im left with 9.8-5.7 but that doesnt give me 4.8 m/s*2 , it gives me 4.1
     
  8. Oct 3, 2009 #7
    hi there again,
    T - f = Ma*a __________ 1
    Mb*g - T = Mb*a ______ 2
    Add 1 and 2, you get: Mb*g - f = (Ma + Mb)a
    Plug in everything except a, and you get exactly 4.775
    Plug in the value of a in any of the two eq's.
     
  9. Oct 4, 2009 #8
    WOW I didn't know that. Seriously... these questions look soo hard but then its simple. THANK YOU VERY VERY MUCH. You don't understand how much you've helped me. Bless you.
     
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