Calculate magnitude and direction of mass's acceleration.

AI Thread Summary
A 2kg mass hanging from a spring with a stiffness of 70N/m oscillates vertically under gravity. For part a, when the spring is stretched to 40cm, the force is calculated as -7N, leading to an acceleration of 3.5 m/s² downward when considering gravity. In part b, similar calculations apply for a stretch of 75cm, with adjustments for the new displacement. The discussion emphasizes the importance of including gravitational force in the total force equation, leading to the formula a = (-kx + mg) / m. Overall, the calculations and understanding of forces acting on the mass are confirmed as correct.
bob29
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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,
 
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for both cases you should also consider the acceleration due to gravity which you haven't...
 
Not exactly sure on the formula but would it be Ftotal= mg + -kx ?
where g=-9.8ms^-2.
 
bob29 said:
Not exactly sure on the formula but would it be Ftotal= mg + -kx ?
where g=-9.8ms^-2.

yes the force is right.. go ahead! :smile:
 
from there would it be correct if i were to let Ftotal=ma?
then ma = -kx + mg, so that a= (-kx + mg) / m
 
yes again :smile:
 

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