Question

hen a 0.20-kg block is suspended from a vertically hanging spring, it stretches the spring from its original length of 0.050 m to 0.060 m.The same block is attached to the same spring and placed on a horizontal, frictionless surface. The block is then pulled so that the spring stretches to a total length of 0.10 m. The block is released at time t= 0 and undergoes simple harmonic motion. Page 24.What is the maximum acceleration ofthe block?

Answer 1

Answer

mass of the block = 0.2 Kg

it stretches the spring from its original length of 0.050 m to 0.060 m.

total length = 0.1 m

FOR , VERTICAL SYSTEM:

K Δx = mg ,    Δx = 0.01 m

$k = \dfrac{mg}{\Delta x}$

$k = \dfrac{0.2 \times 9.8}{0.01}$

k =196 N/m

$\omega = \sqrt{\dfrac{k}{m}}$

$\omega = \sqrt{\dfrac{196}{0.2}}$

$\omega = 31.305\ rad/s$

$f = \dfrac{\omega}{2\pi}$

$f = \dfrac{31.305}{2\pi}$

f = 4.98 Hz

for horizontal system

k Δ x = ma

$a = \dfrac{k \Delta x}{m}$

$a = \dfrac{196\times (0.1-0.05)}{0.2}$

$a = \dfrac{196\times 0.05}{0.2}$

a = 49 m/s²