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A 205-g block is pressed against a spring of force constant 1.18 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 = 3.391m

Explanation:

a) Given:

Mass of the block, m = 205g=0.205kg

spring constant, k = 1.18 kN/m

Angle made by the ramp = 60°

displacement of the spring, x = 10cm= 0.1m

Now,

The initial potential energy of the spring = final gravitational potential energy of block

mathematically,

[tex]\frac{1}{2}kx^2 = mgh[/tex]

where,

g = acceleration due to gravity

h = height of the block above the ground = LsinΘ

L is the displacement of the block parallel to the ramp

[tex]\frac{1}{2}kx^2 = mgLsin\theta[/tex]

substituting the values in the above equation we get

[tex]\frac{1}{2}1.18\times 1000\times 0.1^2 = 0.205\times 9.8\times Lsin60^o[/tex]

or

[tex]5.9 = 1.739\times L[/tex]

or

L = 3.391m

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