Answer:
Explanation:
Given that
F=-αxy²j
α=3N/m³
Displacement from x=2.85 to y=2.85
Since, y=x
Then dy=dx
ds=dx i + dy j
a. Work done
W=∫ F.ds from (0,0) to (2.85,2.85)
W=∫(-αxy²j).(dx i + dy j) (0,0)—(2.85,2.85)
Note, i.i=j.j=1 i.j=j.i=0
W=∫-αxy²dy from (0,0)—(2.85,2.85)
W=-α∫xy²dy from (0,0)—(2.85,2.85)
x=y
W=-α∫y³dy from (0,0)—(2.85,2.85)
W=-αy⁴/4. from (0,0)—(2.85,2.85)
W=-3[(2.85)⁴/4-0]
W=-49.48J
The work done on the tool by the force is -49.48J