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A 180-g block is pressed against a spring of force constant 1.35 kN/m until the block compresses the spring 10.0 cm. The spring rests at the bottom of a ramp inclined at 60.0° to the horizontal. Using energy considerations, determine how far up the incline the block moves from its initial position before it stops under the following conditions.

Respuesta :

Answer:

L = 4.32 m

Explanation:

Here we can use the energy conservation to find the distance that it will move

As per energy conservation we can say that the energy stored in the spring = gravitational potential energy

[tex]\frac{1}{2}kx^2 = mg(L + x)sin\theta[/tex]

[tex]\frac{1}{2}(1.35 \times 10^3)(0.10^2) = (0.180)(9.8)(L + 0.10)sin60[/tex]

now we need to solve above equation for length L

[tex]6.75 = 1.53(L + 0.10)[/tex]

[tex]L + 0.10 = 4.42[/tex]

[tex]L = 4.42 - 0.10[/tex]

[tex]L = 4.32 m[/tex]

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