Question

The kinetic energy of an object attached to a horizontal ideal spring is denoted by KE and the elastic potential energy by PE. For the simple harmonic motion of this object the maximum kinetic energy and the maximum elastic potential energy during an oscillation cycle are KEmax and PEmax, respectively. In the absence of friction, air resistance, and any other nonconservative forces, which of the following equations applies to the object-spring system? A. KE + PE = constant B. KEmax = PEmax

Answer 1

Option A and B

Step-by-step explanation:

In the absence of non-conservative forces, the mechanical energy would remain conserved for any system. Mechanical energy is the sum of kinetic energy and potential energy.

Thus, option A is correct.

K.E. + P.E. = constant

Horizontal spring under goes simple harmonic motion. when the spring would be fully extended or compressed, it would have maximum potential energy and kinetic energy would be zero.

When the spring crosses the equilibrium point, the spring does not have potential energy but has maximum kinetic energy that is maximum speed.

Thus, option B is correct.

`KE_(max) = PE_(max)`

Answer 2

Both A and B

Step-by-step explanation:

Given that

Kinetic energy = KE

Potential energy = PE

For simple harmonic motion

The maximum kinetic energy occurs at the mean position but on the other hand maximum potential energy occurs at the extreme positions .But from the energy conservation ( if there is no any resistance force)

1.

Maximum kinetic energy = Maximum potential energy

KE(max) = PE(max)

2.

kinetic energy + potential energy = Constant

KE + PE = C

So both the option is correct.