Figure shows three forces applied to a block that moves leftward by 3m over a smooth floor. The magnitudes of force are F1=5N,F2=9N and F3=3N. The net work done on the block by the three forces will be
A
1.5J
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
3.6J
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
3J
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
4.5J
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A1.5J
From the FBD of body and considering the rightwards as +vex− direction and upwards as +vey− direction , →F=−5^i+9cos60∘^i+9sin60∘^j−3^j =−5^i+92^i+9√32^j−3^j =−5^i+92^i+9√32^j−3^j →F=−12^i+(9√32−3)^j Displacement of the block →S=−3^im Work done by force, W=→F⋅→S=[−12^i+(9√32−3)^j]⋅(−3^i) ∴W=1.5J