A block of mass m = 3.05 kg is pushed a distance d = 6.85 m...

 

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A block of mass m = 3.05 kg is pushed a distance d = 6.85 m along a frictionless horizontal table by a constant applied force of magnitude F = 16.0 N directed at an angle = 32.0 below the horizontal as shown in the figure below. (a) Determine the work done on the block by the applied force. (b) Determine the work done on the block by the normal force exerted by the table. (c) Determine the work done on the block by the force of gravity. (d) Determine the work done by the net force on the block. Step 1 (a) The work W done on an object by a constant force during a linear displacement A is defined to be W = (Fcose)d where d is the magnitude of the displacement and 9 is the angle between the vectors F and the displacement vector AX. If we redraw the figure with the force acting at the center of the block as in the diagram below, we can more easily see the direction of the applied force F relative to the direction of the displacement AX AX The component of F parallel to A points in the same direction as AX, so that the work is positive, and so that we have where W W FAXcose Fdcos app N(C app is the work done by the applied force F. mcos [ 3, A block of mass m = 3.05 kg is pushed a distance d = 6.85 m along a frictionless horizontal table by a constant applied force of magnitude F = 16.0 N directed at an angle = 32.0 below the horizontal as shown in the figure below. (a) Determine the work done on the block by the applied force. (b) Determine the work done on the block by the normal force exerted by the table. (c) Determine the work done on the block by the force of gravity. (d) Determine the work done by the net force on the block. Step 1 (a) The work W done on an object by a constant force during a linear displacement A is defined to be W = (Fcose)d where d is the magnitude of the displacement and 9 is the angle between the vectors F and the displacement vector AX. If we redraw the figure with the force acting at the center of the block as in the diagram below, we can more easily see the direction of the applied force F relative to the direction of the displacement AX AX The component of F parallel to A points in the same direction as AX, so that the work is positive, and so that we have where W W FAXcose Fdcos app N(C app is the work done by the applied force F. mcos [ 3,