Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Work Power Energy

Question:

The potential energy of a particle is determined by the expression \(U = \alpha (x^2 + y^2)\), where \(\alpha\) is a positive constant. The particle begins to move from a point with coordinates (3, 3), only under the action of potential field force. Then its kinetic energy T at the instant when the particle is at a point with coordinates (1, 1) is : 

Options:

\(zero\)

\(12 \alpha\)

\(8 \alpha\)

\(16 \alpha\)

Correct Answer:

\(16 \alpha\)

Explanation:

As the particle moves only under the action of conservative force, its mechanical energy must be conserved. So :

\(\Delta K + \Delta U = 0\)

\(\Rightarrow \Delta T = - \Delta U = -(U_i - U_f)\)

\(\Rightarrow \Delta T = -[\alpha(1^2+1^2)-\alpha(3^2+3^2)]\)

  \(= 16 \alpha\)