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Spring motion/Friction/Freefall problem

  1. Dec 10, 2011 #1
    1. The problem statement, all variables and given/known data
    A 400 gram block is launched across a horizontal frictionless table by a horizontal spring that has been compressed 32 cm and has a spring constant of 56 N/m. After going a short distance, the block slides across a 30 cm strip at the edge of the table that has a coefficient of friction of 0.52 between the block and the table. The block then flies off the table and lands on the floor. If the table is 1.25 m high, how far from the edge of the table does it land on the floor?


    2. Relevant equations
    motion equations
    f=-kx
    Ff=N(coeff of fric)


    3. The attempt at a solution
    i started off with a 56N/m spring being compressed 32 cm gives a 17.92 N force which translates to an acceleration of 44.8 m/s/s for a .4 kg block. I do not know how to handle the problem from here/ how to deal with the 30 cm strip of friction
     
  2. jcsd
  3. Dec 10, 2011 #2
    well the force of friction is μmg

    you're going to want to use energy as well to find the initial velocity of the block after it leaves the spring.

    (1/2)kx2 = (1/2)mv2 then you need to solve for v to find the initial velocity.

    next you need to find the deceleration of the block due to the friction of the plane.

    which i think is μg because the block has left the spring so no more force is applied to the block.

    so the block is going to travel a distance of 30cm, so knowing this your deceleration and initial velocity you can solve for the final velocity at the end of the table.

    You know the height of the table so you need to find the time it takes for a block to fall that height under the influence of gravity. once you know that just multiply your final velocity at the end of the table by that time and there you have it.
     
  4. Dec 10, 2011 #3
    so using 1/2kx^2 i get (1/2)(56Nm)(.32m)^2 = 2.87 which through (1/2)mv^2 yields a velocity of 3.79 m/s. Then the decleration is equal to μg = (.52)(-9.8) = -5.096 m/s/s. thru motion equation i got final velocity = 3.36m/s. then thru another motion equation i got time = .51sec for the object to fall 1.25m. Then through another motion equation i got .86m as distance traveled, check this please?
     
  5. Dec 10, 2011 #4
    EDIT: I think .86m is wrong, i think i jus ttake 3.36m/s and multiply by .51s and get 1.71m???
     
  6. Dec 10, 2011 #5
    well i would use t=0.505s multiplied by 3.36m/s and yeah that would be right
     
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