Question 333: Work, Energy and Power & Circular Motion

Question

A particle is moved along a path AB-BC-CD-DE-EF-FA, as shown in figure, in presence of a force $$F = (\alpha y\widehat i + 2\alpha x\widehat j)$$ N, where x and y are in meter and $$\alpha $$ = $$ - $$1 Nm
-1
. The work done on the particle by this force F will be ............... Joule.

Accepted Answer

d = F.dr
d = $$\alpha $$ydx + 2$$\alpha $$xdy
A$$ 	o $$B, y = 1, dy = 0
then $${W_{A 	o B}} = \int {\alpha ydx} $$$$ = \alpha 1\int_0^1 {dx}  = \alpha $$
B$$ 	o $$C, x = 1, dx = 0
then $${W_{B 	o C}} = 2\alpha 1\int_1^{0.5} {dy}  =  - 2\alpha (0.5) =  - \alpha $$
C$$ 	o $$D, y = 0.5, dy = 0
then $${W_{C 	o D}} = \int_1^{0.5} {\alpha ydx}  = \alpha .{1 \over 2}\int_1^{0.5} {dx}  =  - {\alpha  \over 4}$$
D$$ 	o $$E, x = 0.5, dx = 0
then $${W_{D 	o E}} = 2\alpha \int {xdy}  = 2\alpha .{1 \over 2}\int\limits_{0.5}^0 {dy}  =  - {\alpha  \over 2}$$
E$$ 	o $$F, y = 0, dy = 0 then W
EF
= 0
F$$ 	o $$A, x = 0, dx = 0 then W
F$$ 	o $$A
= 0
$$ 	herefore $$ $$W = \alpha  - \alpha  - {\alpha  \over 4} - {\alpha  \over 2} =  - {{3\alpha } \over 4}$$
Given, $$\alpha  =  - 1 $$
$$\Rightarrow W = {3 \over 4}J = 0.75J$$

Work, Energy and Power & Circular Motion

A particle is moved along a path AB-BC-CD-DE-EF-FA, as shown in figure, in presence of a force $F = (\alpha y\widehat i + 2\alpha x\widehat j) N, where x and y are in meter and \alpha = - $1 Nm -1 . The work done on the particle by this force F will be ............... Joule.

Integer / Numeric Answer Type