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A 120-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 400 N. For the first 17 m the floor is frictionless, and for the next 17 m the coefficient of friction is 0.34. What is the final speed of the crate?

  • physics - ,

    find the work that went into the system: Force*distance

    Now, find the frictional force on the next portion:
    Then find the work done on friction:

    Finally, find the KE of the crate
    KE crate=workin-frictionwork
    1/2 mv^2=400*34-mu(mg)17

    v^2=800*34/mass -2*mu*g*17
    solve for v
    check my thinking.

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