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Kinematics/Force help - especially with direction

  1. Jun 12, 2012 #1
    1. The problem statement, all variables and given/known data
    1. An object has a mass of 45kg and is pulled along a horizontal surface by a rope that makes an angle 32 above the horizontal. If the coefficient of friction for the surface is 0.13 and the tension on the rope is 95N what is the acceleration of the object along the horizontal? (0.66m/s is the ans in my textbook)

    2. A student is standing 10m from a tall building, if he throws a ball at 12m/s at an angle of 35 up from the horizontal at what height will the ball hit the bulidng (1.92m)

    3. [10kg]----[20kg]----> 100N (applied) [weights in boxes]
    Given the diagram find the following if the surface has a coefficient of friction of 0.30

    Find the accleration of the system (2.06m/s) i keep getting 0.39
    Find the tension connecting the 10kg block to the 20kg block (50N)

    4. If a student throws and object into the air with a speed of 16m/s at an angle of 25 above the horizontal, if the building is 75m tall how far from the base will the object hit the ground

    5. A young woman swims at 2m/s@ 45 S of W for 5 minutes then she swims at 1.5m/s @ 80 E of N for 4 min
    What is the average velocity of the woman (40.94 m/min @ 79 S of W ) i cant seem to find this answer

    It may seem like alot, but ive tried my best at these questions and cant figure them out please even one will help

    @phantom what you see is exactly what i see i can take a picture if you really need the [10kg]--[20kg]---- are boxes being pulled on a rope by 100N applied force
    2. Relevant equations

    f= ma
    v=d/t
    a=v/t
    d=vit+1/2at^2
    Ff=μ|fg|

    3. The attempt at a solution
    1. no idea
    2. no idea

    3. fg+fg+t=ma
    a = fg+fg+t/m
    fg1 = 10 * 9.81 *.3 = 29.43
    fg2= 20 * 9.81 *.3 = 58.86

    100N + (-29.43N) + (-58.86N)/30kg = 0.39m/s
    Tension if i cannot get accerlation right i cant get the tension right so i didnt even try.

    4. no idea
    5. 2m/s * 600s = 1200m
    1.5 * 240s= 360

    (-1200 * cos45 + 360 * cos10)^2 + (-1200*sin45 + 360 * sin10)^2 - than square rooted didnt get the right answer
     
    Last edited: Jun 12, 2012
  2. jcsd
  3. Jun 12, 2012 #2

    PhanthomJay

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    You must show what your attempt was so that we may better assist in determining where you are stuck.

    Also, posting 5 problems in one post is a bit much. It s best to post each problem separately, showing your attempt, in order for you to get adequate help for each.

    You might want to start with problem 3, showing how you arrived at your answer, and better describing the problem as given. Neither your answer nor the book's make sense without a better description and wording of the problem.
     
  4. Jun 13, 2012 #3

    PhanthomJay

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    Problem 3: if the coefficient of kinetic friction is 0.3 for each block, your answer for the acceleration is correct. I don't know where the book answer of 2.06 comes from.
     
  5. Jun 13, 2012 #4
    Question 5:

    You made a mistake in the math, 5 minutes is 300 seconds, not 600 seconds.
     
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