one end of a spring with a force constant of k = 10.0 N/m is attached to the end
ID: 1574683 • Letter: O
Question
one end of a spring with a force constant of k = 10.0 N/m is attached to the end of a long horizontal frictionless track and the other end s attached to a mass m = 2.20 kg which glides along the track. After you establish the equilibrium position of the mass-spring system, you move the mass in the negative direction (to the left), compressing the spring 2.48 m. You then release the mass from rest and start your stopwatch, that is x(t = 0) =-A, and the mass executes simple harmonic motion about the equilibrium position (a) Determine the displacement of the mass (magnitude and direction) 1.0 s after it is released. magnitude direction Select (b) Determine the velocity of the mass (magnitude and direction) 1.0 s after it is released. magnitude direction Select (c) Determine the acceleration of the mass (magnitude and direction) 1.0 s after it is released magnitude direction Select (d) Determine the force the spring exerts on the mass (magnitude and direction) 1.0 s after it is released magnitude direction Select (e) How many times does the object oscillate in 12.0 s? (Do not round your answer to an integer.) oscillationsExplanation / Answer
w = sqrt(k/m) = sqrt(10/2.20) = 2.132 rad/s
A = 2.48 m
x = - A cos(wt)
(A) putting t = 1 sec
x = - 2.48 m cos(2.132 rad)
x = 1.32 m
Ans: 1.32 m, right
(B) v = dx/dt = 5.287 m/s sin(2.132 t)
t = 1 aec
v = 4.48 m/s
Ans: 4.48 m/s, right
(C) a = dv/dt = (11.27 m/s^2 ) cos(2.132 t)
t = 1 s
a = - 6 m/s^2
Ans: 6 m/s^2 , left
(D) F = m a = -13.2 N
Ans: 13.2 N , left
(e) T = 2 pi / w = 2.947 sec
n = 12 / T = 4.07
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