Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति
JEE Main & Advanced
Physics
Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति
Question Bank
Self Evaluation Test - Work, Energy and Power
question_answer
A particle is placed at the origin and a force \[F=kx\] is acting on it (where k is positive constant). If\[U(0)=0\], the graph of \[U(x)\]versus x will be (where U is the potential energy function)
A)
B)
C)
D)
Correct Answer:
A
Solution :
[a] \[U=-\int_{0}^{x}{Fdx}=-\int_{0}^{x}{kxdx=-\frac{1}{2}k{{x}^{2}}}\] It is correctly drawn in [a]