Consider a block-spring system where the equilibrium position of the block is de

Question

Consider a block-spring system where the equilibrium position of the block is defined to be at x=0. The spring constant is k=25 N/m. The block has a initial velocity to the right. There is a frictional force F=3 N between the mass and the horizontal surface (do not distinguish between static and kinetic friction). We define A as the amplitude of the oscillation if there was no frictional force. Select true or false for the following statements. If A is 0.339 m then the block stops at x=0 after reversing its directional x =0.240m. if a is 0.168m, then the block stops after reversing its direction. If A is 0.252m then the block reverses its direction at x = 0.159m before stopping. Lf the initial kinetic energy is 10.0 J, then A is equal to 0.894 m.

Explanation / Answer

1)
Initial K.E = 1/2 * k x^2 = 1/2 * 0.339^2 * 25
K.E = 1.4365 J

Now Energy Lost in Friction = Force * distance
At 0.240 m , Energy Lost to Friction = 3 * 0.240 = 0.72 J
Energy Converted to P.E = 1/2 * k * x^2 = 1/2 * 25 * 0.240^2 = 0.72 J

True.

2)
False.
3)
Inital K.E = 1/2 * m*v^2 = 1/2 * k * x^2
K.E = 1/2 * 25 * 0.252^2 = 0.7938 J

At 0.159 m , Energy Lost to Friction = 3 * 0.159 = 0.477 J
Energy Converted to P.E = 1/2 * k * x^2 = 1/2 * 25 * 0.159^2 = 0.316 J

Total Energy = 0.316 + 0.477 = 0.793 J

True.

4)
Initial K.E = 10.0 J
1/2 * kx^2 = 10.0
x = sqrt(20/25)
x = 0.894 m
True

Correct option - TFTT

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