A block of mass m = 0.53 kg is attached to a spring with force constant 137 N/m

Question

A block of mass m = 0.53 kg is attached to a spring with force constant 137 N/m is free to move on a frictionless, horizontal surface as in the figure below. The block is released from rest after the spring is stretched a distance A = 0.13 m. (Indicate the direction with the sign of your answer. Assume that the positive direction is to the right.)

(a) At that instant, find the force on the block.
N

(b) At that instant, find its acceleration.
m/s2

Explanation / Answer

(a) By F = -kx =>F = -137x 0.13 = -17.81 N [-ve is just indicating the direction of Force] (b) By a(x) = - ?^2x =>a(x) = -(2pn)^2x =>a(x) = -(2p x 1/2pv[k/m])^2x =>a(x) = -(k/m)x =>a(x) = -(137 x 0.13)/0.53 =>a(x) = -33.60 m/s^2 [-ve indicating that it is retardation]

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