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Strike441 [17]
2 years ago
8

A 360.0 $g$ block is dropped onto a vertical spring with a spring constant k = 254.0 $N/m$. The block becomes attached to the sp

ring, and the spring compresses 0.26 $m$ before momentarily stopping. While the spring is being compressed, what work is done by the block's weight?
Physics
1 answer:
tino4ka555 [31]2 years ago
7 0

Answer:

8.6 J

Explanation:

Work done by the block = change in energy in the spring

W = ½ kx²

W = ½ (254.0 N/m) (0.26 m)²

W = 8.6 J

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A) The resultant force is 30.4 N at 25.3^{\circ}

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Explanation:

A)

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B)

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\theta=180^{\circ}-60^{\circ}=120^{\circ}

So we have:

F_{2x}=F_2 cos \theta = (15)(cos 120)=-7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 120)=13.0 N

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And the direction is:

\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{13.5})=43.9^{\circ}

Learn more about vector addition:

brainly.com/question/4945130

brainly.com/question/5892298

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