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.

Answer 1

dw = \(\vec F.​​d\vec r\)

dw = αydx + 2αxdy

\(A\to B\) \(y=1,dy=0\) \(w_{A\to B}=\int \alpha yd\mathrm x,\) \(1\int^1_0d\mathrm x = \alpha\)

\(B\to C\) \(\mathrm x=1, d\mathrm x = 0 \) \(w_{B\to C}=2\alpha.1\int^{0.5}_1dy\) \(=-2\alpha(0.5)=-\alpha\)

\(C\to D\) \(y=0.5,dy=0\) \(w_{C\to D}=\int^{0.5}_1\alpha yd\mathrm x = \alpha.\frac{1}{2}\), \(\int^{0.5}_1d\mathrm x=-\frac{\alpha}{4}\)

\(D\to E\) \(\mathrm x=0.5,d\mathrm x = 0\) \(w_{D\to E}=2\alpha \int \mathrm xdy = 2\alpha.\frac{1}{2}\) \(\int^0_{0.5}dy = -\frac{\alpha}{2}\)

\(E \to F,\) \(y=0, E_{EF}=0\)

\(F\to A,\mathrm x = 0, d\mathrm x = 0, W_{F\to A}=0\)

\(\therefore w=\alpha-\alpha-\frac{\alpha}{4}-\frac{\alpha}{2}\) \(=-\frac{3\alpha}{4}\)

Given α =-1

=>W=+3/4J

=0.75J.

Answer 2

|F|= αyi+2αxj

α = -1 

∴ F - [yi + 2xj]

\(\int\)\(\vec{F}.\vec{dl}\) = –(1 \(\times\) 1) + \(\left(2\times \frac{1}{2}\right)\) + \(\left(\frac{1}{2}\times \frac{1}{2}\right)\) + \(\left(1\times \frac{1}{2}\right)\)

= +0.75 J

ΔW = +0.75 J

Answer 3

\(F=(\alpha y \hat i + 2\alpha \mathrm x \hat j)\)

\(W_{AB}=(-1\hat i).(1\hat i)\) = –1J

\(\begin{bmatrix}\vec F = -1\hat i + 2\alpha \mathrm x \hat j\\\vec S = 1\hat i\end{bmatrix}\)

Similarly,

W_BC = 1J

W_CD = 0.25J 

W_DE = 0.5 J 

W_EF = W_FA = 0 J

New work in cycle = 0.75 J

Answer 4

Given: the path is AB-BC-CD-DE-EF-FA

force F = (αyi+2αxj) N

The work done on the particle by this force F will be

W_AB = \(\int\limits^1_0\alpha yd\mathrm x\) = –1

W_BC = \(\int\limits^{0.5}_12\alpha \mathrm xd y\) = +1

W_CD = \(\int\limits^{0.5}_1\alpha yd \mathrm x \) = +0.25

W_DE = \(\int\limits^0_{0.5}2\alpha \mathrm xdy \) = +0.5

W_EF = W_FA = 0

W_net = 0.75