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NEET
Exam
NEET
Subject
Physics
Chapter
Work, Power and Energy

NEET · Physics · Work, Power and Energy

A particle is placed at the origin and a force F = kx is acting on...

A particle is placed at the origin and a force F=kxF = kx is acting on it (where kk is a positive constant). IF U(0)=0U ( 0 ) = 0, the graph of U(x)U ( x ) versus xx will be (where UU is the potential energy function)A particle is placed at the origin and a force F = kx is acting on it (where k is a positive constant). IF U ( 0 ) = 0A particle is placed at the origin and a force F = kx is acting on it (where k is a positive constant). IF U ( 0 ) = 0A particle is placed at the origin and a force F = kx is acting on it (where k is a positive constant). IF U ( 0 ) = 0A particle is placed at the origin and a force F = kx is acting on it (where k is a positive constant). IF U ( 0 ) = 0A particle is placed at the origin and a force F = kx is acting on it (where k is a positive constant). IF U ( 0 ) = 0

Options

  1. AIf the largest possible value of x such that 0 < x < π satisfying the equation (2 2 x+1)/( x+ x)= 2 is (a π)/(b) . Find
  2. BIf the largest possible value of x such that 0 < x < π satisfying the equation (2 2 x+1)/( x+ x)= 2 is (a π)/(b) . Find
  3. CIf the largest possible value of x such that 0 < x < π satisfying the equation (2 2 x+1)/( x+ x)= 2 is (a π)/(b) . Find
  4. DIf the largest possible value of x such that 0 < x < π satisfying the equation (2 2 x+1)/( x+ x)= 2 is (a π)/(b) . Find

Correct answer

A. If the largest possible value of x such that 0 < x < π satisfying the equation (2 2 x+1)/( x+ x)= 2 is (a π)/(b) . Find

Step-by-step solution

F=−dUdx⇒∫dU=−∫Fdx\text{F} = - \dfrac{\text{dU}}{\text{dx}} ⇒ ∫ \text{dU} = - ∫ \text{Fdx}

U2−U1=−[∫kxdx]⇒U=−12kx2{\text{U}}_{2} - {\text{U}}_{1} = - \left[∫ \text{kxdx}\right] ⇒ \text{U} = - \dfrac{1}{2} {\text{kx}}^{2}

A particle is placed at the origin and a force F = kx is acting on it (where k is a positive constant). IF U ( 0 ) = 0
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