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Aforce of 120 n is applied to a 50-kg mass in the direction of motion for a distance of 7.0 m and then the force is increased to 160 n for the next 3.0 m. for the 10 m of travel, how much work is done by the varying force? select one: a. 1 320 j b. 66 000 j c. 26.4 j d. ​1 220 j e. 1 480 j

Answers

  • Réponse publiée par: elaineeee

    The answer is a. 1320 J.

    Explanation:

    The problem is asking for the total work done by the varying force. Work is the product of force and the displacement caused by the force exerted (W=Fd) . To get this, we must use the equation:

    W = F_{1} d_{1} +F_{2} d_{2}

    In this particular problem,

    F_{1} = 120 N

    d_{1} = 7.0 m

    F_{2} = 160 N

    d_{2} = 3.0 m

    Substituting these values, we get

    W = F_{1} d_{1} + F_{2} d_{2}

    W = (120 N) (7.0 m) +  (160 N) (3.0 m)

    W = 1320 J

    Note that 1 Nm = 1 J

    I hope this helps. :)

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Aforce of 120 n is applied to a 50-kg mass in the direction of motion for a distance of 7.0 m and th...