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

A body of mass 5 kg starts from rest and moves in a straight line under the influence of two variable forces F1=(10t+6t) N and F2=(156t) N, acting in same direction. Find the power delivered by force F1 at t=2 sec.

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

The correct option is B 230 J/s
Total force F=F1+F2
F=(10t+6t)+(156t)
F=10t+15
ma=10t=15
5a=10t+15
a=2t+3
dvdt=2t+3

Integrating both sides:
v0dv=20(2t+3)dt
v=(t2+3t)|20
v=2×2+3×2=10 m/s

Work done by force F1
=F1.v=F1vcosθ
Here θ=0
Work done by force F1 at t=2 s
=(10×2+62)×10=230 J/s

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon