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NEET
Exam
NEET
Subject
Physics
Chapter
Work, Power and Energy

NEET · Physics · Work, Power and Energy

A particle is moved along a path AB - BC - CD - DE - EF - FA, as...

A particle is moved along a path AB−BC−CD−DE−EF−FA,AB - BC - CD - DE - EF - FA, as shown in figure, in presence of a force F→=\overset{\to}{F} =\left(\alphay\overset{^}{i} + 2\alphax\overset{^}{j}\right)N,wherewherex and yy are in meter and α=−1N/m−1.\alpha = - 1N / {m}^{- 1} . The work done on the particle by this force F→\overset{\to}{F} will be ____ Joule.
A particle is moved along a path AB - BC - CD - DE - EF - FA, as shown in figure, in presence of a force → F = ( ^ i +

Options

  1. A0.75
  2. B0.25
  3. C1.25
  4. D1.75

Correct answer

A. 0.75

Step-by-step solution

dw=F→⋅dr→dw = \overset{\to}{F} ⋅ d\overset{\to}{r}
dw=aydx+2axdydw = aydx + 2axdy
Now, total work done in whole path is given by,
W=WAB+WBC+WCD+WDE+WEF+WFAW = {W}_{AB} + {W}_{BC} + {W}_{CD} + {W}_{DE} + {W}_{EF} + {W}_{FA}
A→B,y=1,dy=0,WA→B=∫αydx=α⋅1∫01dx=αA \to B, y = 1, dy = 0, {W}_{A \to B} = ∫ \alpha ydx = \alpha ⋅ 1\mathop{∫}_{0}^{1}dx = \alpha
B→C,x=1,dx=0,WB→C=2α⋅1∫10.5dy=−2α(0.5)=−αB \to C, x = 1, dx = 0, {W}_{B \to C} = 2\alpha ⋅ 1\mathop{∫}_{1}^{0.5}dy = - 2\alpha\left(0.5\right) = - \alpha
C→D,y=0.5,dy=0,WC→D=∫10.5αydx=α⋅12∫10.5dx=−α4C \to D, y = 0.5, dy = 0, {W}_{C \to D} = \mathop{∫}_{1}^{0.5}\alpha ydx =\alpha ⋅ \dfrac{1}{2}\mathop{∫}_{1}^{0.5}dx = - \dfrac{\alpha}{4}
D→E,x=0.5,dx=0,WD→E=2α∫xdy=2α⋅12∫10.5dy=−α2D \to E, x = 0.5, dx = 0, {W}_{D \to E} = 2\alpha ∫ xdy = 2\alpha ⋅ \dfrac{1}{2}\mathop{∫}_{1}^{0.5}dy = - \dfrac{\alpha}{2}
E→F,y=0,dy=0,WEF=0E \to F, y = 0, dy = 0, {W}_{EF} = 0
F→A,x=0,dx=0,WE→A=0F \to A, x = 0, dx = 0, {W}_{E \to A} = 0
∴W=α−α−α4−α2=−3α4∴ W = \alpha - \alpha - \dfrac{\alpha}{4} - \dfrac{\alpha}{2} = - \dfrac{3\alpha}{4}
Given α=−1⇒W=+34J=0.75J\alpha = - 1 ⇒ W = + \dfrac{3}{4}J = 0.75J
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