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Ksju [112]
3 years ago
5

A spring with spring constant of 26 N/m is stretched 0.22 m from its equilibrium position. How much work must be done to stretch

it an additional 0.12 m
Physics
1 answer:
Evgen [1.6K]3 years ago
8 0

Answer:

1.503 J

Explanation:

Work done in stretching a spring = 1/2ke²

W = 1/2ke²........................... Equation 1

Where W = work done, k = spring constant, e = extension.

Given: k = 26 N/m, e = (0.22+0.12), = 0.34 m.

Substitute into equation 1

W = 1/2(26)(0.34²)

W = 13(0.1156)

W = 1.503 J.

Hence the work done to stretch it an additional 0.12 m = 1.503 J

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A person throws a pumpkin at a horizontal speed of 4.0 — off a cliff. The pumpkin travels 9.5 m horizontally
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Complete Question

A person throws a pumpkin at a horizontal speed of   4.0 m/s off a cliff. The pumpkin travels 9.5m horizontally before it hits the ground. We can ignore air resistance.What is the pumpkin's vertical displacement during the throw? What is the pumpkin's vertical velocity when it hits the ground?

Answer:

The  pumpkin's vertical displacement  is  H = 27 .67 \ m

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Explanation:

From the question we are told that

   The  horizontal speed is  v_h  =  4 m/s

    The horizontal distance traveled is  d =  9.5 \ m

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Where t is the time taken

substituting values

          9.5 =  4 * t

   =>     t =  \frac{9.5}{4}

            t = 2.38 \ sec

Now the vertical displacement is mathematically represented as

        H  =  v_v t  +  \frac{1}{2} a_v t^2

now the vertical velocity before the throw is  zero

    So

          H =  0 +  \frac{1}{2} (9.8) * (2.375)^2

          H = 27 .67 \ m

Now the final vertical velocity  is mathematically represented as

          v_v__{f}} =  v_v + at

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             v_v__{f}} =  0 + (9.8)* (2.375)

            v_v__{f}} = 23.298 \  m/s

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Answer:

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Answer:

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