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A block is attached to an ideal spring. The spring is laid out horizontally along the x-axis, and the block is on a frictionless surface. When the block is at x = 0, the spring is at its natural length. The block is given an initial kinetic energy of 9.00 J, when the block is at x = 0, with the initial velocity directed in the positive x-direction. That kinetic energy is enough to cause the block to move out to x = +d before reversing direction. How much kinetic energy would the block have to have at x = 0 for the reversal of direction to occur at x = +2d, instead of x = +d? _______ J

User Fraggjkee
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1 Answer

3 votes

Answer:

K = 36 J

Step-by-step explanation:

In this exercise we must use the conservation of mechanical energy at two points at initial x = 0 and the end point of maximum compression of the spring

Initial x = 0

Em₀ = K = ½ m v²

Final point of maximum compression


Em_(f) = Ke = ½ k x²

Em₀ =
Em_(f)

K = ½ k x²

We put the kinetic energy because it is a data, let's look for the spring constant

k = 2 K / x²

k = 2 9 / d²

k = 18 / d²

Let's calculate the kinetic energy, so that the compression is x = 2d

K = ½ k x²

K = ½ (18 / d²) (2d)²

K = ½ 18/d² 4d²

K = 36 J

User Jonboy
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