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. a 1.20 kg box is firmly attached to a spring with a spring constant of 32.0 n/m (the other end of the spring is attached to a wall); the surface the box is on is frictionless surface. you stretch the spring by 0.500 m and hold the box steady for a moment. you then begin pushing it with a constant 4.00 n force directed parallel to the surface (the figure shows the point where the spring is neither stretched nor compressed and the direction of the force you apply). hot compress will the spring will the spring be at the next instant the box is at rest?

Respuesta :

If the spring is pushed with a constant force of 4 N force directed parallel to the surface, the spring will be compressed by 0.5 m at the next instant the box is at rest.

Us = 1 / 2 k x²

W = F x

Us = Potential energy of the spring

k = Spring constant

x = Displacement

F = Force

W = Work done

k = 32 N / m

F = 4 N

xi = 0.5 m

According to law of conservation of energy,

Change in potential energy = Total Work done by the force

1 /2 k xf² - 1 / 2 k xi² = F xf + F xi

1 / 2 * 32 ( xf² - 0.5² ) = 4 ( xf + 0.5 )

4 ( xf² - 0.25 ) = xf + 0.5

4 xf² - xf - 1.5 = 0

This is in the form of a quadratic equation,

x = - b ± √ ( b² - 4 a c ) / 2 a

a = 4

b = - 1

c = - 1.5

xf = - ( - 1 ) ± √ [ ( - 1 )² - ( 4 * 4 * -1.5 ) ] / ( 2 * 4 )

xf = 1 ± √ 25 / 8

xf = 1 ± 5 / 8

xf = 0.75 or - 0.5

Since it is compressed,

xf = - 0.5 m

Therefore, the spring will be compressed by 0.5 m at the next instant the box is at rest.

To know more about Potential energy of the spring

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