How is Work and Energy Calculated in a Horizontal Table Setup?

[Kudo Shinichi]

HELP!Work and energy problem

Homework Statement

A block of mass 1.5kg rests on top of a horizontal table.
a)It is pushed in a straight line a distance of 1.0m in the x-direction, and then in a straight line a distance of 0.5m in the y-direction. If the block is at rest initially and at the end of the displacement, how much work was done by the external agent doing the pushing? The coefficient of kinetic friction between the block and the table is mu(k)=0.25
b)Calculate the work done by the external agent in pushing the block in a straight line from its initial to its final position. Compare with the answer in part a). Is friction a conservative force? explain

The Attempt at a Solution

a) Frictional force=mu(k) times normal force
=-0.25*1.5*9.8
=-3.675 (it is negative because it is the opposite to the applied force)
applied force is equal and opposite to the frictional force
work done in x-direction=3.675*1.0=3.675
work done in y-direction=3.675*05=1.8375
add both works together=5.51J

b) use Pythagorean theorem to find out the straight line, which is the hypotenuse in this case, which is 1.118
1.118*3.675=4.109J
Yes the friction is a conservative force, because we used the same force in part b as in part a and friction is just equal and opposite to the applied force

 

[christensen]

Except for the final line about the conservative force, this is all right.

Friction is not a conservative force. We know this because for it to be conservative the work done on an object to move it from point "a" to point "b" will be the same regardless of the path it takes. However, you proved this to be incorrect for friction as it takes MORE energy to go the 1m up and 0.5m right, instead of the 1.118 diagonally.

 

[Kudo Shinichi]

Except for the final line about the conservative force, this is all right.

Friction is not a conservative force. We know this because for it to be conservative the work done on an object to move it from point "a" to point "b" will be the same regardless of the path it takes. However, you proved this to be incorrect for friction as it takes MORE energy to go the 1m up and 0.5m right, instead of the 1.118 diagonally.

thank you very much