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physics

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A person pushes a 11.3-kg shopping cart at a constant velocity for a distance of 19.4 m on a flat horizontal surface. She pushes in a direction 29.2 ° below the horizontal. A 44.6-N frictional force opposes the motion of the cart. (a) What is the magnitude of the force that the shopper exerts?

  • physics - ,

    Given: m=11.3kg. A = 29.2 deg., Ff = 44.6N.

    Wc = mg = 11.3kg * 9.8N/kg = 110.74N. =
    Weight of cart.
    Fc = (110.74N,0deg.).
    Fp = 110.74sin(0) = 0 = Force parallel to surface.
    Fv = 110.74cos(0) = 110.74N. = Force
    perpendicular to surface.

    Ff = 44.6N = Force of fiction.

    Fn=Fap*cos29.2 - Fp - Ff = ma = 0, a=0.
    Fap*cos29.2 - 0 - 44.6 = 0,
    Fap*cos29.2 = 44.6,
    Fap = 44.6 / cos29.2 = 51.1N. = Force
    applied.

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