Physics

A 25.0kg pickle is accelerated from rest through a distance of 6.0m in 4.0s across a level floor. If the friction force between the pickle and the floor is 3.8N, how much work is required to move the object?

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asked by Sandara
  1. d = Vo*t + 0.5a*t^2 = 6 m.
    0 + 0.5a*4^2 = 6
    8a = 6
    a = 0.75 m/s^2.

    Fp = m*g = 25kg * 9.8N/kg = 245 N. = Force of pickle.

    !@#$%^&-Fp-Fk = m*a
    !@#$%^&-245-3.8 = 25*0.75
    !@#$%^& = 18.75+245+3.8 = 267.6 N. = Force
    applied.

    Work = !@#$%^& * d = 267.6 * 6 = 1605 Joules

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    posted by Henry
  2. d = Vo*t + 0.5a*t^2 = 6 m.
    0 + 0.5a*4^2 = 6
    8a = 6
    a = 0.75 m/s^2.

    Fnet=m*a
    Fnet=25*0.75=18.75 N

    Fnet=!@#$%^&-Ff
    !@#$%^&=Fnet+Ff
    !@#$%^&=18.75+3.8=22.55 N

    W=!@#$%^&*d
    W=22.55*6=135.3 J

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    posted by Marco

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