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Ipatiy [6.2K]
2 years ago
13

A 0.250 kg block attached to a light spring oscillates on a frictionless, horizontal table. The oscillation amplitude is A = 0.1

25 m and the block moves at 3.00 m/s as it passes through equilibrium at x = 0. (a) Find the spring constant, k.
Physics
1 answer:
bulgar [2K]2 years ago
8 0

The spring constant, k will be 18 N/m.The ratio of force to one unit of displaced length is known as the spring constant.

<h3>What is the spring constant?</h3>

Spring constant is defined as the ratio of force per unit displaced length.

Given data;

Mass of  block,m=0.250 k

Amplitude,A = 0.125 m

Velocity,V=3.00 m/

Spring constant,K=?

The formula for the spring constant when the spring is oscillate at the given amplitude A:

\rm K = \frac{V_{max}^2 m}{A} \\\\  K = \frac{3.0 \times 0.250}{0.125} \\\\ K=18 \ N/m

Hence, the spring constant, k will be 18 N/m

To learn more about the spring constant, refer:

brainly.com/question/4291098

#SPJ1

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<h2>Part (a)</h2>

We are given/can infer these variables:

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We want to find the displacement and the final velocity of the rock.

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We can use this equation to find the final velocity:

  • v = v_0 + at

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The final velocity of the rock is -39.2 m/s.

Now we can use this equation to find the displacement of the rock:

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Plug in the known variables into this equation.

  • Δx = 0 * 4.0 + 1/2(-9.8)(4.0)²
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  • Δx = -78.4 m

The displacement of the rock is -78.4 m.

<h2>Part (b)</h2>

We are given/can infer these variables:

  • v_0 = 8.0 m/s
  • a = -9.8 m/s²
  • t = 4.0 s

We can use this equation to find the final velocity:

  • v = v_0 + at

Plug in the known variables into this equation.

  • v = 8.0 + (-9.8)(4.0)
  • v = 8.0 + -39.2
  • v = -31.2 m/s

The final velocity of the rock is -31.2 m/s.

We can use this equation to find the displacement:

  • Δx = v_0 t + 1/2at²

Plug in known variables:

  • Δx = 8.0(4.0) + 1/2(-9.8)(4.0)²
  • Δx = 32 - 4.9(16)
  • Δx = -46.4 m

The displacement of the rock is -46.4 m.

<h2>Part (c)</h2>

We are given/can infer these variables:

  • v_0 = -8.0 m/s
  • a = -9.8 m/s²
  • t = 4.0 s

We can use this equation to find the final velocity:

  • v = v_0 + at

Plug in the known variables into this equation.

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The final velocity of the rock is -47.2 m/s.

We can use this equation to find the displacement:

  • Δx = v_0 t + 1/2at²

Plug in known variables:

  • Δx = -8.0(4.0) + 1/2(-9.8)(4.0)²
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The displacement of the rock is -110.4 m.

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