Homework/Assignment Help

Question

when a 2.50-kg object is hung vertically on a certain lightspring described by hooke’s law, the spring stretches 2.76 cm

a)what is the force constant of the spring ?

b)if the 2.50-kg object is removed ,how far will the springstretch if a 1.25-kg block is hung on it?

c) how much work must an external agent do to stretch the samespring 8.00 cm from its unstretched position?

Answer

a)

F=kΔx

k = F/Δx = 2.50kg*9.8m/s² /2.76*10^-2m = 887.68N/m

b)

F=kΔx

Δx = F/k = 1.25kg*9.8m/s²/ 887.68N/m=0.0138m

c)

the work done must equal to the potential energy stored by thespring after the stecher

w = (1/2)k(Δx)² = (1/2)*887.68N/m*(8.00*10^-2m)² =2.84J

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