wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A force F=4x^i+2y^j displaces a body from r1=3^i+8^j to r2=9^i6^j. Find the work done by the force.

A
125 J
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
120 J
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
116 J
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
Zero
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 116 J
Gives force =F=4x^i+2y^j
initial position r1=3^i+8^j
and portion r2=9^i6^j
We know work done
W=s2s1F.ds
=s2s1F.(dx^i+dy^j)
W=s2s1(4x^i+2y^j).(dx^i+dy^j)
=934xdx+682ydy
W=[2x2]93+[y2]68
W=144+(28)
W=116 J

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
The Work Done as a Dot-Product
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon