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object of mass m=2.6 kg is dropped from rest at a height of h above a massless spring with spring constant k=494.3 N/m that is initially at its equilibrium position. The block falls onto the spring and compresses it a distance d=0.66 m below its equilibrium position before momentarily bringing the block to a stop. Find the height h from which the block was dropped (above the equilibrium point of the spring).

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MechEngineer

Answer:

3.56 m

Explanation:

As the object falls from height h, and then compresses the spring a distance of 0.66m, its initial potential energy is converted into elastics energy of the spring.  

Let g = 9.81m/s2. Knowing that the total change in potential height is h + d, we have the following equation

[tex]E_p = E_s[/tex]

[tex]mg(h + d) = kd^2/2[/tex]

[tex]2.6*9.81(h + 0.66) = 494.3*0.66^2/2[/tex]

[tex]h + 0.66 = 4.22[/tex]

[tex]h = 4.22 - 0.66 = 3.56 m[/tex]

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