Question

A conservative force acts on a 4kg particle in X
direction. The potential energy U(X) is given by
U(x) = 40 + (x - 6)^2, where x is in metre. If it is
given that at x = 8 m, KE is 30 J, then find
maximum KE of the particle.
(1) 10J
(2) 17 J
(3) 24 J
(4) 34 J​

Answer 1

Answer:

Option (4) 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

$U=40$ J

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.

Answer 2

Answer:

See the attachment

Good luck

Explanation: