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22 June, 03:23

You have a light spring which obeys Hooke's law. This spring stretches 2.66 cm vertically when a 2.70 kg object is suspended from it. Determine the following. (a) the force constant of the spring (in N/m).

(b) the distance (in cm) the spring stretches if you replace the 2.70 kg object with a 1.35 kg object.

(c) the amount of work (in J) an external agent must do to stretch the spring 8.40 cm from its unstretched position.

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  1. 22 June, 03:33
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    a) k = 994.74 N/m

    b) Δx = 0.0133 m

    c) W = 7.02 J

    Explanation:

    a) let Δd = 2.66*10^-2 m and the forces acting on the object at 2.66 cm are the force by spring Fs and force by gravity Fg such that:

    Fs = Fg

    k*Δd = m*g

    k = (m*g) / (Δd)

    k = (2.70*9.8) / (2.66*10^-2)

    k = 994.74 N/m

    b) let Δx be the distance the spring stretches and the forces acting on the object at Δx are the force by spring Fs and force by gravity Fg such that:

    Fs = Fg

    k*Δx = m*g

    Δx = (m*g) / (k)

    = (1.35*9.8) / (994.74)

    = 0.0133 m

    c) when the spring stretch X = 8.40 cm the only forces are Fs = force by spring and F = force by external agent such that:

    F = Fs

    = k*X

    = (994.74) * (8.40*10^-2)

    = 83.56 N

    then the work done by the external agent is given by:

    W = F*X

    = (83.56) * (8.40*10^-2)

    = 7.02 J
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