wiz-icon
MyQuestionIcon
MyQuestionIcon
3
You visited us 3 times! Enjoying our articles? Unlock Full Access!
Question

A particle of mass 5 kg is initially at rest. A force starts acting on it along a fixed direction. The magnitude of the force changes with time as shown in the figure. Find the velocity of the particle at the end of 20 s.


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

The correct option is D 20 m/s
Impulse imparted by force is given by J=F.dt
i.e Impulse = area under the Ft curve
J=(5×2)+(5×4)+(5×6)+(5×8)
J=100 N-s

Initial velocity v1=0 m/s
particle is at rest initially.
Let the velocity of the particle at the end of 20 s be v2
Impulse imparted J=m Δv
J=m(v2v1)
100=5(v20)
v2=20 m/s

flag
Suggest Corrections
thumbs-up
2
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon