A block of mass m = 3.60 kg is pushed a distance d = 2.60 m along a frictionless, horizontal table by a constant applied force of magnitude F = 11.0 N directed at an angle è = 28.0° below the horizontal as shown in the figure below.
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To find the work done on the block, we need to calculate the dot product between the applied force and the displacement of the block.
The dot product of two vectors A and B is given by:
A · B = |A| |B| cos(θ)
In this case, the force vector F is given by F = 11.0 N at an angle of 28.0° below the horizontal. We can break down this force into its horizontal and vertical components.
The horizontal component Fx is given by:
Fx = F cos(θ)
Fx = 11.0 N * cos(28.0°)
The vertical component Fy is given by:
Fy = F sin(θ)
Fy = 11.0 N * sin(28.0°)
Since the block is being pushed horizontally along the table without any vertical displacement, the vertical component of the force (Fy) does not contribute to the work done.
Therefore, the work done on the block is given by:
Work = Fx * d
Substituting the given values:
Work = (11.0 N * cos(28.0°)) * 2.60 m
Calculating this expression will give us the work done on the block.