Given below is the plot of a potential energy function U(X) for a system, in which a particle is in one dimenstional motion, while a conservative force F(X) acts on it. Suppose that Emech=8J, the incorrect statement for this system is :
A
at X>X4, K.E. is constant throughout the region.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
at X<X1, K.E. is smallest and the particle is moving at the slowest speed.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
at X=X2, K.E. is greatest and the particle is moving at the fastest speed.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
at X=X3, K.E =4J.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B at X<X1, K.E. is smallest and the particle is moving at the slowest speed.
We have from the given diagram Emech=8J
(A) at X>X4, U=constant=6J K.E=Emech−U=2J=constant
(B) at X<X1, U=constant=8J K.E=Emech−U=8−8=0J
Particle is at rest.
(C) At X=X2U=0⇒Emech=K.E=8J K.E is greatest, and particle is moving at fastest speed.
(D) At (X=X3),U=constant=4J K.E=Emech−U=(8−4)J=4J
Hence, option (b) is correct.