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Question:

A particle is moved along a path AB−BC−CD−DE−EF−FA, as shown in figure in presence of a force F→=(αyi^+2αxj^)N, where x and y are in meter and α=−1;Nm−1;. The work done on the particle by this force F→ will be _______ joule.

0.75

Solution:

Correct option is A. 0.75
dw=F→.d→
dw=αydx+2αxdy
A→B y=1, dy=0
wA→B=∫αydx=α,1∫01dx=α
B→C x=1, dx=0
wB→C=2α.1∫10.5dy=−2;α(0.5)=−α
C→D y=0.5 dy=0
wC→D=∫10.5αydx=α.12,∫10.5dx=−α4
D→E x=0.5 dx=0
wD→E=2α∫xdy=2α.12∫0.50dy=−α2
E→F, y=0, dy=0, WEF=0
F→A, x=0, dx=0, WF→A=0
∴w=α−α−α4−α2=−3α4
Given α=−1; ⇒W=+3/4J=0.75.