a mass m on a horizontal surface with mild kinetic friction mu k is attached to

Question

a mass m on a horizontal surface with mild kinetic friction mu k is attached to an ideal spring with spring constant k. when the spring is compressed by a distance d to the left of its equilibrium position and released, the mass starts to slide back to the right. ignore any static friction, assume kinetic fruition is weak enough that mass accelerates to the right.

x= -d x=0

express final answers in only terms of k, m, mu k, and variable x

(a) find an expression for the acceleration of the mass as a function of its position x as the mass slides back to its quilibrium position between x= -d and x=0

(b) use your answer from part a, find the position x between -d and 0 where the mass's acceleration momentarily equals zero

Explanation / Answer

at any position between -d to 0 this relation will be true-

1/2k(d-x)^2=1/2mv^2 ( loss in potential energy =gain in kinetic energy)

differentiating both sides with respect to time (t) , we get-

-k(d-x)v=mva, where a is acceleratin and v is velocity,

therefore a=-k(d-x)/m

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