Three forces →F1=(2^i+4^j)N;→F2=(2^j−^k)N and →F3=(^k−4^i−2^j)N are applied on an object of mass 1kg at rest at origin. The position of the object at t=2s will be
A
(−2m,−6m)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(−4m,8m)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
(3m,6m)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(2m,−3m)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B(−4m,8m) From Newton's second law, Fnet=→F1+→F2+→F3=m→anet ⇒2i+4j+2j−k+k−4i−2j=1→a ⇒→a=−2i+4j ⇒ Now →s=→ut+12→at2 as body is initially at rest →s=12→at2 →s=12(−2i+4j)t2 →s=−4i+8j m So (−4m,8m)