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    A block of ice of mass 5kg is sliding down a plane inclined at 35 degrees to the horizontal. A man pushes up on the block parrelel to the incline so that its speed changes from 0.90 m/s to 0.33 m/s in sliding 3.5m down the plane.
    (i) if the coefficient of friction between ice and incline is 0.10, what is the magnitude of the force exerted on the block by the man?

    I work out the force of kinetic friction Fk = (0.10)(5.0)(-9.8)cos35 = -4.01 N

    Total work done parallel to incline = 0.5mvf^2 - 0.5mvi^2 = 1.75J

    W = fxdcos35 = 0.61N

    Force exerted on ice due to man: F(bm) = -4.01--0.610 = 4.62N up the incline

    (ii) Find the work done by each of the four forces that act on the block. What is the net work done on the block by the man?
    Please help, Im a bit confused, maybe mistake above?

    F1 = Normal = F2 = Weight =
    F3 = Friction = -4.01 N
    F4 = Man = 4.62 N

    Net work sum of these?
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    (Original post by Everybody)
    A block of ice of mass 5kg is sliding down a plane inclined at 35 degrees to the horizontal. A man pushes up on the block parrelel to the incline so that its speed changes from 0.90 m/s to 0.33 m/s in sliding 3.5m down the plane.
    (i) if the coefficient of friction between ice and incline is 0.10, what is the magnitude of the force exerted on the block by the man?

    I work out the force of kinetic friction Fk = (0.10)(5.0)(-9.8)cos35 = -4.01 N

    Total work done parallel to incline = 0.5mvf^2 - 0.5mvi^2 = 1.75J

    W = fxdcos35 = 0.61N
    I can't understand this equation. What are you calculating here?

    Were you trying to calculate the net force acting on the ice? Then there is a problem with your notation, since you wrote that work is equal -4.62N. I suggest F=W/d. You do not need to include the cos35 function, because the distance is measured parallel to the inclined plane.


    Force exerted on ice due to man: F(bm) = -4.01--0.610 = 4.62N up the incline
    Total work done is, as you correctly calculated, -1.75J. There are three forces that do this work: frictional force (which does negative work), gravitational force (which does positive work) and the man's force. You seem to have neglected the gravitational force, and as pointed out above -0.61N is wrong.


    (ii) Find the work done by each of the four forces that act on the block. What is the net work done on the block by the man?
    Please help, Im a bit confused, maybe mistake above?

    F1 = Normal = F2 = Weight =
    F3 = Friction = -4.01 N
    F4 = Man = 4.62 N
    Forces are correct, with the reservation that F_1\neq F_2. Normal force is a reaction force - it reacts to the component of weight of the block that is perpendicular to the slope. It ensures that the block does not "sink" into the plane.

    As for the work done by each of the forces, remember W_i=\mathbf{F}_i\mathbf{d}\cos \theta? Try to find magnitudes of all forces and the angles between each force and the direction of the movement of the block of ice.
 
 
 
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