Question

-/7 POINTS SERCP9 5.P.008.MI.SA. MY NOTES ASK YOUR TEAC This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise A block of mass m 1.95 kg is pushed a distance d = 5.15 m along a frictionless horizontal table by a constant applied force of magnitude F-16.0 N directed at an angle - 31.0° below the horizontal as shown In the figure below. () 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 F during a linear displacement AR is defined to be W = (Fcosed whered is the magnitude of the displacement and is the angle between the vectors F and the displacement vector A. 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 relative to the direction of the displacement AR AR The component of parallel to All points in the same direction as AR, so that the work is positive, and so that we have Wapp - Faxcos - Fdcose ( N m cos where Wapp is the work done by the applied force F. Submit Skip you can come back)

Answer 1

(a) The work done by applied force is, W = Fd cose =(16) (5.15) cos(31.0°) = 70.63 J (b) The work done by normal force is, W = 0 (c) The work done by gravitational force is, W=0 The total work done on the block is, Wotal = 70.63 J The work done is equal to scalar product of force and displacement is, Wapp = F Arcos = Fd cosa =((F)N)((d)m) (cosa)' = (Fd cose)J