Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

A force F=−K(yi+xj)  (where K is a positive constant) acts on a particle moving in the xy-plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0) and then parallel to the y-axis to the point (a, a). The total work done by the force F on the particles is

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya

a

−2Ka2

b

2Ka2

c

-Ka2

d

Ka2

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

While moving from (0,0) to (a,0) Along positive x-axis, y = 0  ∴ F→=−kxj^i.e.  force is in negative y-direction while displacement is in positive x-direction.∴W1=0Because force is perpendicular to displacement Then particle moves from (a, 0) to (a, a) along a line parallel to y-axis (x=+a) during this  F→=−k(yi^+aJ^)The first component of force, −kyi^ will not contribute any work because this component is along negative    x-direction (−i^) while displacement is in positive           y-direction (a,0) to (a,a). The second component of force i.e. −kaj^ will perform negative work∴W2=(−kaj^) (aj^)=(−ka) (a)​ =−ka2So net work done on the particle  W=W1+W2=0+(−ka2)=−ka2
Watch 3-min video & get full concept clarity

courses

No courses found

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon