A mass m1 = 5.5 kg rests on a frictionless table and connected by a massless string to another mass m2 = 4.1 kg.?

A mass m1 = 5.5 kg rests on a frictionless table and connected by a massless string to another mass m2 = 4.1 kg. A force of magnitude F = 26 N pulls m1 to the left a distance d = 0.83 m.

1)How much work is done by the force F on the two block system?J 2)How much work is done by the normal force on m1 and m2?J 3)What is the final speed of the two blocks?m/s 4)How much work is done by the tension (in-between the blocks) on block m2?J 5)What is the tension in the string?N 6)The net work done by all the forces acting on m1 is:positivezeronegative7)What is the NET work done on m1?

2 Answers

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  • NCS
    Lv 7
    4 months ago
    Favorite Answer

    1) W = F*d = 26N * 0.83m = 22 J

    2) That force is perpendicular to the displacement, so the work done is 0

    3) total mass is 9.6 kg, so

    KE = ½mv²

    22 J = ½ * 9.6kg * v²

    means that the final velocity is

    v = 2.1 m/s

    4) W2 = ½ * 4.1kg * (2.1m/s)² = 9.2 J

    5) T = W2 / d = 9.2J / 0.83m = 11 N

    6) positive (since its KE increases)

    7) W1 = ½ * 5.5kg * (2.1m/s)² = 12 J

    All answers to 2 significant digits. Hope this helps!

  • 4 months ago

    The answer to last one is 12.45J

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