Physics Problem Help??

The ball launcher in a pinball machine has a spring that has a force constant of 1.20N/cm . the surface on which the ball moves is inclined 10.0 degrees with respect to the horizontal. If the spring is initially compressed 5cm find the launching speed of 100g ball when the plunger is released. Friction and the mass of the plunger are negligible.

The answer should be 1.68 m/s...... How i dont know!!!, and this Q is coming n Exam and im still working on it.

Thx For helping .

2 Answers

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  • 1 decade ago
    Favorite Answer

    It's all about conservation of energy. When the spring is compressed 5 cm it has a potential energy of kx²/2 where k is the constant and x is the compression (and they're in the MKS system, not the CGS system ☺) so

    Ep = 120*(.05)²/2 = .15J That's *all* the energy there is in the system. To push a 100g ball a height of

    .05*sin(10) = 0.008682 m requires

    mgh = .1*9.8*.008 = .00784J of that energy. The remainder is available to be converted into speed (kinetic energy) so .15-.008682 = 0.141318J and since Ek = mv²/2

    v = √(2*.141318/.1) = 1.681178 rounded to 1.68 m/s.

    Doug

  • 1 decade ago

    Hi,

    This problem is easy if you only think interms of energy:

    Stored Potential Energy Of The Spring(SE)=

    Final Potential Energy Of The Ball when launched(PE)

    + Final Kinetic Energy Of Ball(KE)

    .........{Conservation Of Energy}

    So ,the stored Potential Energy of the spring is

    = 1/2*k*x^2

    =0.15J .....{k=1.2N/cm , x = 5cm;remember to convert

    cm to m}

    And, the final Potential energy of the ball is when it leaves the plunger,which means you need the vertical component of the 5cm length = 0.05sin10 m

    So the PE = m*g*h

    = 0.0085 J (taking g=9.83)

    And the final K.E. = 1/2*m*vel^2

    = 0.05*v^2

    Now all you have to do is replpace the values:

    0.15 = 0.0085 + 0.05*v^2

    v^2 = 2.83

    On solving, v = 1.68 m/s

    Best Of Luck

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