A horizontal spring attached to a wall has a force constant of k = 900 N/m. A bl

Question

A horizontal spring attached to a wall has a force constant of k = 900 N/m. A block of mass m = 1.10 kg is attached to the spring and rests on a horizontal, frictionless surface as in the figure below.

(a) The block is pulled to a position xi = 6.80 cm from equilibrium and released. Find the elastic potential energy stored in the spring when the block is 6.80 cm from equilibrium and when the block passes through equilibrium.

(b) Find the speed of the block as it passes through the equilibrium point.


(c) What is the speed of the block when it is at a position xi/2 = 3.4 cm?


(d) Why isn't the answer to part (c) half the answer to part (b)?

Explanation / Answer

elastic potential energy PEi = 0.5*k*xi^2

Ui = 0.5*900*0.068^2 = 2.0808 J

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b)

speed at xi is vi = 0

kinetic enegy at xi = 0.5*m*vi^2 = 0

total energy at xi = KEi + PEi = 0.5*k*xi^2

PE at x =0 is U = 0

let speed at x = 0 is vb

KE at x = 0 is KE = 0.5*m*vb^2

total energy at x = 0 is 0.5*m*vb^2

from energy conservation

energy at xi = energy at at (x=0)

0.5*k*x^2 = 0.5*m*vb^2

vb =sqrt(k/m)*xi

vb = sqrt(900/1.1)*0.068 = 1.945 m/s <-------answer

part(c)

speed at x = xi/2 = vc

PE at x = xi/2 = 0.5*k*xi^2/4

KE at x = xi/2 = 0.5*m*vc^2

total energy = 0.5*k*xi^2/4 + 0.5*m*vc^2

from energy conservation

energy at xi = energy at at (x=xi/2)

0.5*k*xi^2 = 0.5*k*xi^2/4 + 0.5*m*vc^2

vc = sqrt(3k/4m)*xi

vc = sqrt(3*900/4*1.1)*0.068 = 1.68 m/s <-------answer

part(d)

here part of kinetic converted to potential energy

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