Question

Minimum PE of a particle of mass 2kg performing SHM is -10J .If maximum Velocity of particle is 2m/s.Find Maximum potential Energy during the motion.​

Answer 1

$\huge{\underline{\sf{\red{Answer\hookrightarrow-6J}}}}$

Given:

Mass = 2kg

Minimum Potential Energy= -10J

Maximum Velocity = 2m/s

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Formula

KE$_max$=$\dfrac{1}{2} mv^2$

M= Momentum,K=Kinetic Energy,U=Potential Energy

M$_i$=M$_f$ $\implies$$K_{min}+ U_{max}=K_{max} +U_{min}$

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$\huge{\underline{\sf{\blue{Solution}}}}$

KE = $\dfrac{1}{2} (2)(2)^2$

$\implies$ $\large{\boxed{4J}}}$

Since the momentum is conserved

$\implies$ Initial and Final Momentum Are Equal

By above formula

$\implies$$K_{min}+ U_{max}=K_{max} +U_{min}$

$\implies$ $0+U_{max} = 4+(-10)$

$\implies$ $\large{\boxed{\boxed{-6J}}}$

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