Calculate magnitude and direction of mass's acceleration.

[bob29]

Homework Statement

Question: "A block of 2kg mass is hanging from a spring with stiffness K=70N/m. The spring has an unstretched length of 30cm. The mass oscillates vertically under the influence of the spring and gravity. Calculate the magnitude and direction of the mass's acceleration when:
a) x=40cm
b) x=75cm

Homework Equations

F = -kx
x = (L' - L) where L = unstretched
F = ma
F = mg

The Attempt at a Solution

part a) F= -KX, F= -70*(0.4-0.3) = -7N, F=ma where F= -7N so a=m/F = (-3.5)j N,
direction of A is downward @ a magnitude of (3.5+9.8)ms^-2

part b) same as part a with the number change.

Not sure if it's correct, would like to know where I went wrong.
Cheers,

 

[The legend]

for both cases you should also consider the acceleration due to gravity which you haven't...

 

[bob29]

Not exactly sure on the formula but would it be F_total= mg + -kx ?
where g=-9.8ms^-2.

 

[The legend]

Not exactly sure on the formula but would it be F_total= mg + -kx ?
where g=-9.8ms^-2.

yes the force is right.. go ahead!

 

[bob29]

from there would it be correct if i were to let F_total=ma?
then ma = -kx + mg, so that a= (-kx + mg) / m

 

[The legend]

yes again