Item 5 Part A A toy of mass 0.190-kg is u horizontal spring with force constant

Question

Item 5 Part A A toy of mass 0.190-kg is u horizontal spring with force constant k = 250 N/m . When the toy is a distance 0.0130 m from its equilibrium position, it is observed to have a speed of 0.200 m/s SHM on the end of a What is the toy's total energy at any point of its motion? Express your answer with the appropriate units. EValue Units Submit My Answers Give UR Part B What is the toy's amplitude of the motion? Express your answer with the appropriate units. AValue Units Submit My Answers Give Up Part C What is the toy's maximum speed during its motion? Express your answer with the appropriate units. o,max=1 Value Units Submit My Answers Give Up Continue

Explanation / Answer

(B) angular velocity, = (k/m) = (250N/m / 0.190kg) .. .. = 36.27rad/s

SHM velocity ,v = (r² - x²) .. . x = displacement, r = amplitude (radius)
(v/)² = r² - x²
r² = (v/)² + x²
r² = (0.200m/s / 36.27rad/s)² + (0.0130m)² = 1.99*10^-4
Amplitude r = 1.99*10^-4) .. .. r = 1.41*10^-2 m

(A) SHM total energy, E = ½m(r)²
Et = ½ x 0.190kg x (36.27 x 1.41*10^-2)² .. .. Et = 2.48*10^-2 J

(C) v(max) occurs at displacement x = 0 (equilibrium position)
v = (r² - x²) v(max) = r .. .. r = amplitude
v(max) = (36.27rad/s x 1.41*10^-2 m) .. .. v(max) = 0.51 m/s

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