A block with a mass of m = 2.2 kg is lifted slowly from ground level at y = 0 m

Question

A block with a mass of m = 2.2 kg is lifted slowly from ground level at y = 0 m to a height of y = 1.94 m. (a) Calculate the work done against gravity to lift it, and (b) the total work done against gravity to lift it and then to move it horizontally by 1.36 m. (A) Find the work done to lift the block. Analyze Since the block's motion is not accelerated the net vertical force on it is zero. Therefore the applied force lifting the block is equal in magnitude and opposite in direction to the gravitational force, mg. The applied force is in the direction of the displacement, so the work done to lift the block is W = F?y = mg?y = (2.2 kg)(9.80665 m/s2)(1.94 m) = 41.83 J (B) Find the total work done to displace the block vertically and then horizontally. The block is slowly displaced horizontally by 1.36 m. The applied force supporting the block is perpendicular to the displacement, so the work done to move the block horizontally is _____________????? J. Finally, the total work done to first lift the block and then move it horizontally is ________ ??? J. Finalize What happens if, instead, the block is slowly moved downward through a displacement ?y = -1.94 m? Is the work done by the force supporting the block positive or negative? What happened to the energy lost from the system by doing work of that sign?

Explanation / Answer

a) work done against the gravity by man is mgh=2.2*9.8*1.94=41.82J

block is perpendicular to the displacement, so the work done to move the block horizontally is 0 since Work =Fdcos90=0(since cos90=0)

Total Work Done is 41.82J

work done tomoved downward through a displacement ?y = -1.94 m will be2.2*9.8*1.94 cos 180=-41.82J since it is opposite of the force acting by gravity therefore its negetive

Energy is positve during this case

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