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? asked in Science & MathematicsPhysics · 2 years ago

Need help on a physics question?

When playing shuffleboard, a player exerts a constant force of 2.1 N on an initially stationary puck, at an angle 55° below the horizontal. If the player pushes the puck for 1.5 m, how fast is the puck moving when it is released? The mass of a puck is 0.49 kg. Ignore the force of friction. show work

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  • oldschool
    Lv 7
    2 years ago
    Favorite Answer

    ∫F∙dx = Work = 2.1*cos55 *1.5 = ½mV² = ½*0.49*V²

    (2.1*cos55 *1.5)/(½*0.49) = V² = 2.7²

    OR F/m = a = (2.1*cos55)/0.49 = 2.46m/s²

    Vf² = 0² + 2*2.46*1.5 = 2.7²

    Vf = ~ 2.7m/s

  • Elyse Rose
    Lv 7
    2 years ago

    You expect us to give a puck?

  • billrussell42
    Lv 7
    2 years ago

    The force that is in the direction of motion is 2.1 cos 55 = 1.20 N

    F = ma

    a = F/m = 1.2/0.49 = 2.46 m/s²

    v = √(2ad) = √(2•2.46•1.5) = 2.72 m/s

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