A block of mass m = 0.546 kg is fastened to an unstrained horizontal spring whos

Question

A block of mass m = 0.546 kg is fastened to an unstrained horizontal spring whose spring constant is k = 97.8 N/m. The block is given a displacement of +0.142 m, where the + sign indicates that the displacement is along the +x axis, and then released from rest. (a) What is the force (with sign) that the spring exerts on the block just before the block is released? (b) Find the angular frequency of the resulting oscillatory motion. (c) What is the maximum speed of the block? (d) Determine the magnitude of the maximum acceleration of the block.

Explanation / Answer

here,

spring constant , k = 97.8 N/m
mass of block, m = 0.576 kg
displacement, x = 0.142 m

PART A:
Spring force = spring contant * displacement
F = k*x
F = 97.8 * 0.142
F = 13.88 N in positive x axis

PART B:
Angular frequecy = sqrt(k/m)
w = sqrt(97.8/0.576)
w = 13.03 rad/sec

PART C:
By the law of energy conservation :

KE(max) of block = PE(max) of the spring
1/2 * m*v^2 = 1/2*k*x^2

solving for velocity, v
v = sqrt(kx^2/m)
v = sqrt((97.8 * (0.142)^2)/0.576)
v = 1.850 m/s

PART D:
From newton second law we have :
F = ma

solving for acceleration, a
a = F/m
a = 13.88 / 0.576
a = 24.09 m/s^2

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