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A block lies on a horizontal frictionless surface and
isattached to the free end of a spring, with a spring constant of
60N/m. Initially, the spring is at its relaxed length and the
blockis stationary at position x = 0. Then an applied force with
aconstant magnitude of 3 N pulls the block in the positive
directionof the x axis, stretching the spring until the block
stops. Assumethat the stopping point is reached. What is the
position of theblock?

Respuesta :

boffeemadrid

Answer:

0.1 m

Explanation:

F = Force exerted on spring = 3 N

k = Spring constant = 60 N/m

x = Displacement of the block

As the energy of the system is conserved we have

[tex]Fx=\dfrac{1}{2}kx^2[/tex]

[tex]\\\Rightarrow x=\dfrac{2F}{k}[/tex]

[tex]\\\Rightarrow x=\dfrac{2\times 3}{60}[/tex]

[tex]\\\Rightarrow x=0.1\ m[/tex]

The position of the block is 0.1 from the initial position.

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