A 2.6 kg particle moves along an x axis, being propelled by a variable force dir

Question

A 2.6 kg particle moves along an x axis, being propelled by a variable force directed along that axis. Its position is given by x = 3.0 m + (4.0 m/s)t + ct2 - (2.6 m/s3)t3, with x in meters and t in seconds. The factor c is a constant. At t = 3.0 s, the force on the particle has a magnitude of 36 N and is in the negative direction of the axis. What is c in m/s2?  (Note: This question is very similar to a textbook question (Fundamentals of Physics, ISBN 9780470044735, Chapter 5, Problem 9).  Thank you.

Explanation / Answer

mass = 2.6kg x = 3 + 4t + ct2 - 2.6t3 m v = dx/dt = 4 + 2ct - 7.8t2 a = dv/dt = 2c - 15.6t force = mass*acceleration = 2.6(2c - 15.6t) = 5.2c - 40.56t at t = 3s F = -36N => -36 = 5.2c - 40.56*3 = 5.2c - 121.68 => 5.2c = 85.68 => c = 85.68/5.2 = 16.477 m/s2

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