12. +-1 points SerCP1113.P.021. My Notes Ask Your Te A horizontal spring attache

Question

12. +-1 points SerCP1113.P.021. My Notes Ask Your Te A horizontal spring attached to a wall has a force constant of k = 860 N/m. A block of mass m = 1.50 kg is attached to the spring and rests on a frictionless, horizontal surface as in the figure below (a) The block is pulled to a position Xi = 6.20 cm from equilibrium and released. Find the potential energy stored in the spring when the block is 6.20 cm from equilibrium. (b) Find the speed of the block as it passes through the equilibrium position. m/s (c) what is the speed of the block when it is at a position x/2 cm? 3.10 m/s Need Help? Read It

Explanation / Answer

position of the spring with respect to time

equation of motion

x = A*cos(wt)

w = angular speed

t = time

velocity v = rate of change in position

v = dx/dt

v = -A*sinwt*w

v = -A*w*sinwt

v = -A*W*sqrt(1-(coswt)^2)

v = -A*w*sqrt(1-(x/A)^2)

v = -A*w*sqrt(A^2-x^2)/A

v = -w*sqrt(A^2-x^2)

kinetic energy KE = (1/2)*m*v^2 = (1/2)*m*(A^2-v^2)

potential energy PE = (1/2)*K*x^2

spring constatn K = m*w^2

total energy E = KE + PE = (1/2)*K*A^2 = (1/2)*m*w^2*A^2

(a)

positon xi = 6.2 cm = 0.062 m

potential energy Ui = (1/2)*K*xi^2 = (1/2)*860*0.062^2 = 1.65 J <<<------ANSWER

(b)

as the block passes through equilibrium position

x = 0

speed v = w*(sqrt(A^2-x^2)

w = sqrt(k/m)

A = amplitude = maximum displacement = xi = 0.062 m

speed v = sqrt(860/1.5)*sqrt(0.062^2-0^2) = 1.48 m/s <<<------ANSWER

===================

part(c)

given x = xi/2 = 0.031 m

speed v = sqrt(860/1.5)*sqrt(0.062^2-0.031^2) = 1.28 m/s <<<----------ANSWER

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