A block of mass m = 0.775 kg is fastened to an unstrained horizontal spring whos

Question

A block of mass m = 0.775 kg is fastened to an unstrained horizontal spring whose spring constant is k = 75.5 N/m. The block is given a displacement of +0.120 m, where the + sign indicates that the displacement is along the +x axis, and then released from rest. (a) What is the force (magnitude and direction) that the spring exerts on the block just before the block is released? N (b) Find the angular frequency ? of the resulting oscillary motion. rad/s (c) What is the maximum speed of the block? m/s (d) Determine the magnitude of the maximum acceleration of the block. m/s2

Explanation / Answer

(a) By F = -kx

=>F = - 75.5 x 0.120 = -9.06 N [-ve indicating the force is towards origin]

(b) By T = 2 x pi x sqrt[m/k]

=>T = 2 x 3.14 x sqrt[0.775/75.5]

=>T = 0.628 sec

=>omega = (2 x pi)/T = 6.28/0.628 = 10 rad/sec

(c) By the law of energy conservation:-

=> KE(max) of block = PE(max) of the spring

=>1/2 x mv^2 = 1/2kx^2

=>v = sqrt[kx^2/m]

=>v = sqrt[{75.5 x (0.120)^2}/0.775]

=>v = 1.184 m/s

(d) By F = ma

=>a(max) = F(max)/m = 9.06/0.775 = 11.69 m/s^2

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