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2 years ago

15

A spring has a natural length of 28.0 cm. If a 23.0-N force is required to keep it stretched to a length of 36.0 cm, how much wo

rk W is required to stretch it from 28.0 cm to 32.0 cm

1 answer:

krek1111 [17]2 years ago

4 0

Answer:

0.23 J

Explanation:

k*(36 - 28) = 23

so k = 23/8 N/cm

W = k(32 - 28)²/2 = 23/8 * 4²/2 = 23 N-cm = 0.23 J

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D₂= 167,21 cm : Magnitude of the second displacement

β= 21.8° , countercockwise from the positive x-axis: Direction of the second displacement

Explanation:

We find the x-y components for the given vectors:

i: unit vector in x direction

j:unit vector in y direction

D₁: Displacement Vector 1

D₂: Displacement Vector 2

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D₂y=142.83-80.65=62.18 cm

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$D_{2} =\sqrt{(D_{x})^{2} +(D_{y} )^{2} }$

$D_{2} =\sqrt{(155.22)^{2} +(62.18 )^{2} }$

D₂= 167.21 cm

Direction of the second displacement

$\beta = tan^{-1} \frac{D_{y}}{D_{x} }$

$\beta = tan^{-1} \frac{62.18}{155.22 }$

β= 21.8°

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β= 21.8.° , countercockwise from the positive x-axis: Direction of the second displacement

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