My Notes O Ask Your Teache 11. 0.25/1 points I Previous Answers SerCP11 13.P.018

Question

My Notes O Ask Your Teache 11. 0.25/1 points I Previous Answers SerCP11 13.P.018 A 0.35-kg object connected to a light spring with a force constant of 22.6 N/m oscillates on a frictionless horizontal surface. The spring is compressed 4.0 cm and released from rest. (a) Determine the maximum speed of the object. m/s (b) Determine the speed of the object when the spring is compressed 1.5 cm. Your response differs from the correct answer by more than 10%. Double check your calculations. ms (c) Determine the speed of the object as it passes the point 1.5 cm from the equilibrium position. Your response differs from the correct answer by more than 10%. Double check your calculations. m/s (d) For what value of x does the speed equal one-half the maximum speed? Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m Need Help? Rosd n

Explanation / Answer

Given

mass m = 0.35 kg

force constant of the spring is k = 22.6 N/m

compression in the spring is x = 4.0 cm

as we know that when the spring compressed , the restoring force develops in the spring , resulting the elastic potential energy develops in the spring

a) the maximum speed would be at equilibrium point that is the total elastic potential energy converts in to kinetic energy

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

0.35*v^2 = 22.6*0.04^2

solving for v = 0.3214 m/s

b) speed of the object when sping is compressed 1.5 cm

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

0.35*v^2 = 22.6*0.015^2

solving for v = 0.120 m/s

c) speed of the object at it passes the point 1.5 cm is  

conservation of energy  

0.5*k*x^2 = 0.5*m*v^2+0.5*k*x1^1

k*x^2 = m*v^2+k*x1^2

m*v^2 = k(x^2-x1^2)

v^2 = k(x^2-x1^2)/m

v = sqrt(k(x^2-x1^2)/m)

v = sqrt(22.6(0.04^2-0.015^2)/0.35) m/s

v = 0.298 m/s

d) the maximum speed of the object isv = 0.3214 m/s

half of the maximum speed of the object is V/2 = 0.3214/2 = 0.1607 m/s

that is 0.5*k*x^2 = 0.5*0.35*0.1607^2 m/s

0.5*22.6*x^2 = 0.5*0.35*0.1607^2

solving for x

x = 1.2 cm

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