Question

A spring (k = 200 N/m) is suspended with its upper end supported from a ceiling. With the spring hanging in its equilibrium configuration, an object (mass = 2.0 kg) is attached to the lower end and released from rest. What is the speed of the object after it has fallen 4.0 cm?

Answer 1

Answer:

V = 6.81$\frac{m}{s}$

Explanation:

Data:

k = 200 N/m

mass = 2.0 kg

x = 4.0 cm => 0.04m

There will only be vertical force, using 2nd Newton's law:

Fe - P = -m$a_{y}$

the minus sign we use to indicate that the object is going down

Using the Hooke's law:

Fe = k*x      where x is the spring compression

k*x - m*g = m$a_{y}$

$a_{y}$ = $\frac{k*x - m*g}{m}$

$a_{y}$ = $\frac{200*0.04 - 2*9.8}{2}$

$a_{y}$ = 5.8$\frac{m}{s^{2} }$

$V^{2} = Vo^{2} + 2*a_{y} *(X - Xo)$

Vo = o$\frac{m}{s}$

Xo = 0m

$V^{2} = 0 + 2*5.8 *(4)$

$V^{2} = 46.4$

V = 6.81$\frac{m}{s}$