He moves his sandpaper back and forth 45 times over a rusty area, each time moving a total distance of 0.15 m. Kevin pushes the sandpaper ag

Question

He moves his sandpaper back and forth 45 times over a rusty area, each time moving a total distance of 0.15 m. Kevin pushes the sandpaper against the surface with a normal force of 1.8 N. The coefficient of friction for the metal/sandpaper interface is 0.92. How much work is done by the kinetic frictional force during the sanding process

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Thu Giang 6 months 2021-08-19T19:22:28+00:00 1 Answers 0 views 0

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    2021-08-19T19:23:43+00:00

    Answer:

    W = -11.178 J

    Explanation:

    In this question, Work is defined by the equation:

    W = F•dx•cos(θ)

    Where;

    F is force

    x is distance over which the force acts on the object,

    θ = angle between force and direction of travel.

    Now, because the force is constant in this case, we don’t need make the equation an integral expression. Also, since the force of friction is always precisely opposite the direction of travel (i.e θ = 180°), the equation can be rewritten like this:

    W = F•x•cos(180)

    W = -F•x

    The force of friction will be given by the equation:

    F_fric = F_norm × coefficient of friction

    Also, the total work will be the sum of all 45 passes by the sandpaper. Therefore, our final equation, when F_fric is substituted into the work equation, is:

    W = -(45)×(F_norm)×(coeff of friction)×(distance)

    We are given;

    normal force = 1.8 N

    Coefficient of friction for the metal/sandpaper interface = 0.92

    Distance = 0.15 m

    W = -(45) × (1.8) × (0.92) × (0.15)

    W = -11.178 J

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