Marm asked in Science & MathematicsPhysics · 9 years ago

A vertical spring (ignore its mass), whose spring stiffness constant is 900 N/m, is attached to a table and is?

A vertical spring (ignore its mass), whose spring stiffness constant is 900 N/m, is attached to a table and is compressed down 0.100 m

a) What upward speed can it give to a 0.200 kg ball when released?

b) How high above its original position (spring compressed) will the ball fly?

Thank you!!!

Update:

Thanks so much Dr. Zorro! Awesome explanation. I got it! I'll best answer you once they let me =)

2 Answers

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  • 9 years ago
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    The potential energy of the compressed spring isV = 1/2 k x^2.

    This potential energy all is converted into mechanical energy of the ball (since we assume the mass of the spring is zero, we do not have to consider gravitational potential energy of the spring nor kinetic energy of the spring).

    The mechanical energy of the ball at the equilibrium position of the spring consists of:

    a gravitational potential energy (in rising a distance x): M g x

    a kinetic energy: what is left for it: 1/2 k x^2 - M g x

    So for

    a) 1/2 M v^2 = 1/2 k x^2 - M g x

    v = sqrt(k/M x^2 - 2 g x)

    This gives v=6.56 m/s

    b) kinetic energy is zero in starting position and in max height of flight, so potential energies are equal. Relative to the starting position this means

    M g h = 1/2 k x^2, so

    h = k x^2/(2 M g)

    This gives h=2.29 m

  • Anonymous
    4 years ago

    k = 980 s = .13 m = .4 Potential Spring = Kinetic Ball Potential Spring = .5*k*s^2 = .5*980*.13^2 = 8.28 Joules Kinetic Ball = 8.28 Joules = .5*m*v^2 = .5*.4*v^2 A) v = sqrt( 8.28/(.5*.4)) = 6.43 m/s Kinetic Ball = Potential Ball .5*m*v^2 = m*g*h .5*.4*6.43^2 = .4*9.81*h B) h = (.5*.4*6.43^2 )/(.4*9.81) = 2.1 m

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