A 10-kg block is attached to one end of a horizontal spring on a level, frictionless surface. The other end of the spring is attached to a vertical support. The spring obeys Hooke's law and has a spring constant of k = 160 N/m. A physics student pulls the block outward so that the spring stretches by 40 cm. The student releases the block at time t = 0.00 s. Which of the following equations properly gives the position as a function of time? a. x(t) = (- 6.4 m/s2)cos(4.0t) O b. x(t) = (0.40 m)cos(4.0t) O c. x(t) = (4.0 m)cos(0.40t) O d. x(t) = (10 m)cos(4.0t) O e. x(t) = (- 1.6 m/s)sin(4.0t)

Physics for Scientists and Engineers: Foundations and Connections
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Author:Katz, Debora M.
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Chapter16: Oscillations
Section: Chapter Questions
Problem 11PQ: A 1.50-kg mass is attached to a spring with spring constant 33.0 N/m on a frictionless, horizontal...
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A 10-kg block is attached to one end of a horizontal spring on a level, frictionless surface. The other end of the spring is attached to a vertical
support. The spring obeys Hooke's law and has a spring constant of k = 160 N/m. A physics student pulls the block outward so that the spring
stretches by 40 cm. The student releases the block at timet = 0.00 s. Which of the following equations properly gives the position as a function of
time?
O a. x(t) = (- 6.4 m/s-)cos(4.0t)
O b. x(t) = (0.40 m)cos(4.0t)
O c. x(t) = (4.0 m)cos(0.40t)
O d. x(t) = (10 m)cos(4.0t)
O e. x(t) = (- 1.6 m/s)sin(4.0t)
Transcribed Image Text:A 10-kg block is attached to one end of a horizontal spring on a level, frictionless surface. The other end of the spring is attached to a vertical support. The spring obeys Hooke's law and has a spring constant of k = 160 N/m. A physics student pulls the block outward so that the spring stretches by 40 cm. The student releases the block at timet = 0.00 s. Which of the following equations properly gives the position as a function of time? O a. x(t) = (- 6.4 m/s-)cos(4.0t) O b. x(t) = (0.40 m)cos(4.0t) O c. x(t) = (4.0 m)cos(0.40t) O d. x(t) = (10 m)cos(4.0t) O e. x(t) = (- 1.6 m/s)sin(4.0t)
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