Question

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.

Answer 1

Answer:

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