Question

Assuming the incline to be frictionless and the zero of gravitational potential energy to be at the elevation of the horizontal line, Group of answer choices the kinetic energy of the block when it has fully compressed the spring will be the kinetic energy of the block just before it collides with the spring will be equal to mgh. the kinetic energy of the block when it has fully compressed the spring will be equal to mgh. the kinetic energy of the block when it has fully compressed the spring will be zero. the kinetic energy of the block just before it collides with the spring will be kx2. Not saved Questions AnsweredQuestion 1 AnsweredQuestion 2 AnsweredQuestion 3 AnsweredQuestion 4 AnsweredQuestion 5 AnsweredQuestion 6 AnsweredQuestion 7 AnsweredQuestion 8 AnsweredQuestion 9 Haven't Answered YetQuestion 10 Time Elapsed: Hide Attempt due: Oct 12 at 11:59pm

Answer 1

The energy in the system is given by the law of conservation of energy.

The energy stored in the spring plus the gravitational potential energy of the block when it has fully compressed the spring will be equal to m·g·h.

Reason:

The given parameters are;

Surface of the inclined plane = Frictionless

Potential energy at the horizontal line = Zero of gravitational potential

By the law of Conservation of Energy, we have;

The spring will be compressed by a distance, x

The energy stored in the compressed spring, K.E. = (1/2)·k·x²

Energy in the block when the block comes to rest at a height, h₁, will be, P.E. = m·g·h₁

Therefore, by conservation of energy, we have;

The initial potential of the block = The stored energy in the compressed string + The gravitational potential energy of the block when it has compressed.

Therefore, the correct option is; The energy stored in the spring plus the gravitational potential energy of the block when it has fully compressed the spring will be equal to m·g·h

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