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

A force acts on a 30 gm particle in such a way that the position of the particle as a function of time is given by x=3t4t2+t3, where x is in metres and t is in seconds. The work done during the first 4 seconds is


A

5.28 J

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B

450 mJ

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

490 mJ

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

530 mJ

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A

5.28 J


v=dxdt=38t+3t2

v0=3m/s and v4=19m/s

W=12m(v24v20) (According to work energy theorem)

=12×0.03×(19232)=5.28 J


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Conservative Forces and Potential Energy
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon