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problem reference 10.1
a 0.450 kg block is attached to an unstrained horizontal spring with spring constant 74.0 n/m. the block is given a displacement of 0.0800 m and then released from rest.
what is the magnitude of the maximum acceleration of the block?
3.41 m/s²
9.80 m/s²
6.55 m/s²
13.1 m/s²

Explanation:

Step1: Recall Hooke's Law and Newton's 2nd Law

Maximum force: $F_{max} = kx_{max}$; $F=ma$ so $a_{max}=\frac{kx_{max}}{m}$

Step2: Substitute values

$k=74.0\,\text{N/m}$, $x_{max}=0.0800\,\text{m}$, $m=0.450\,\text{kg}$
$a_{max}=\frac{74.0 \times 0.0800}{0.450} = \frac{5.92}{0.450} \approx 13.1\,\text{m/s}^2$

Answer:

13.1 m/s²