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

A force is inclined at 60 to the horizontal produces a displacement of s=(^i+^j) m. If the component of force in horizontal direction is 50 N, the net work done by the force on the body is

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

The correct option is D 137 J
Given, Horizontal component of force (Fx)=50 N
Displacement of the body (s)=(^i+^j) m

Since, Fx=Fcosθ=50 N and θ=60
F=50cos60=100 N

Vertical component of force (Fy)=Fsinθ=100sin60=503 N
So, F=(Fx^i+Fy^j)=(50^i+503^j)
Net work done (W)=F.s
W=(50^i+503^j).(^i+^j)
W=50+503, as 31.73
W=137 J

Hence option D is the correct answer

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