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physics

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You push a 2.0 kg block against a horizontal spring, compressing the spring by 15 cm. Then you release the block, and the spring sends it sliding across a tabletop. It stops 75 cm from where you released it. The spring constant is 200 N/m. What is the block-table coefficient of kinetic friction?

  • physics - ,

    I want the answer. Where is it ?

  • physics - ,

    Let K = Kinetic Energy
    U = Potential Energy
    Wf = Work done by friction
    Ug = Gravitational Potential energy
    Ue = Elastic Potential Energy
    Law of conservation of energy states that:

    K1 + U1 + Wf= K2 + U2

    Given:
    mass = 2 kg
    x = 0.15 m
    k= 200
    d = 0.75 m

    Solution:
    K1 = 0 (because it is not moving)
    U1 = Ug + Ue (Ug =0) (Ue = (1/2)(k)(x^2) )
    K2 = 0
    U2 = 0

    0 + 0 + (1/2)(200)(-0.15)^2 + Wf = 0 + 0
    Wf = -2.25 J

    Observe this. Since the block is moving to the right, the force of the friction is acting to the left.

    Wf = (f)(d)cos(theta)
    -2.25 = friction (0.75)(cos180)
    Force of the friction = 3 Joules

    To find the coefficient of friction, find the normal force acting on the block
    Summation of forces along Y-axis = 0
    Normal force - Weight = 0
    Normal force = (2 kg)(9.8 m/s^2)= 19.6 N

    Friction = (coefficent) (Normal force)
    3 Joules = (coefficient) (19.6 N)

    Coefficient of Kinetic Friction = 0.15306

    :D

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