A block with mass m-1.841 kg is attached to a fixed wall by two springs connecte

Question

A block with mass m-1.841 kg is attached to a fixed wall by two springs connected to each other with spring constants k1 = 17.9 N/m and k2 = 34.4 N/m. The coefficient of static friction between the ground and the block is 0.143. What is the maximum distance you can stretch the block from the equilibrium spring length so that the static friction force keeps the block in place? Express your answer in metres This is a Challenge Problem. Diagram not to scale. Void where prohibited by law. Residents of Quebec must complete a skill-based problem before claiming their prize. Note,I swapped in a new problem here since I was unable to find a solution to the old one that didn't require either energy or calculus k2

Explanation / Answer

Given , k1 = 17.9 N/m

k2 = 34.4 N/m

= 0.143

m = 1.841 kg

Since they are connected in series ,

Equivalent spring constant (keq) = k1*k2/(k1 + k2)

keq = 17.9*34.4/(17.9+34.4)

keq = 11.774 N/m

If the block is stretched from equilibrium position ,

Condition for the static friction to keep the block in place

Spring force = frictional force

Keq*x = *m*g ( here x is the maximum displacement)

11.774*x = 0.143*1.841*9.8

x = 2.58/11.774

x = 0.22 m

Therefore,

maximum distance from equilibrium position = 0.22 m

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