Question

A conservative force acts on a 4 kg partidadirection. The potential energy U(x) isU(X) = 40 + (x - 62, where x is in metregiven that at x = 8 m, KE is 30 J, then findmaximum KE of the particle.​

Answer 1

Answer:

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Answer 2

Answer:

Maximum KE of the particle will be 34 J

Explanation:

Given the expression of $U(x)$

$U(x)=40+(x-6)^2$

Total Energy

$E=U+KE$

The total energy will always be the same

Given at x = 8 m, KE is 30 J

Therefore

Total energy at x = 8 m

$E=40+(8-6)^2+30$

$\implies E=40+4+30$

$\implies E=74$ J

The Potential Energy is minimum at x=6 because in the expression there is x-6 square

or, we can solve it by calculus, the PE will be minimum at x = 6

Thus the minimum Potential Energy

$U=40$

When potential energy will be minimum, the kinetic energy will be maximum

The total energy will remain same i.e. 74 J

Thus,

$74=40+KE$

$\implies KE=74-40$

$\implies KE=34$ J

Hope this helps.