Question

A 0.16-kg block on a horizontal frictionless surface is attached to an ideal spring whose force constant (spring constant) is 360 N/m. The block is pulled from its equilibrium position at x = 0.000 m to a position x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. When the position is x = -0.037 m, what is the acceleration of the block? 8

Answer 1

Answer:

$a = 83.25 m/s^2$

Explanation:

As we know that the mean position of the block is given as x = 0

so when block is at x = -0.037 m then compression in the spring is given as

$\Delta x = (x_i - x_f)$

$\Delta x = 0 + 0.037 = 0.037 m$

now we know that spring force is given as

$F = k\Delta x$

$F = 360(0.037)$

$F = 13.32 N$

Now the acceleration of the block is given as

$a = \frac{F}{m}$

$a = \frac{13.32}{0.16}$

$a = 83.25 m/s^2$