Question 28

[Guest]

how did you get that value for x(max)?

[admin]

The passage states that the system was stretched 20 cm or 20 x 10^-2 m from the equilibrium position.

[zeina_ghou6141]

I have another question on #28:

Why isn't weight subtracted at the end? Overall, I was very confused about this problem. Can you explain it in a different way, perhaps? Thank you!

[StoneColdPS14]

the explanation to the question considers the position of the mass at the very bottom of streched spring. which is not the point of maximum force acting on the mass.

the maximum force experienced by the mass attached to the spring will be when it is at the highest point, and the spring is compressed to its maximum. here, you will have Fspring max (Kx) and weight(mg) both acting downward on the mass:

Fnet = K*Xmax + m*g = m*a

K=50N/m
Xmax=20cm=.2m
m=.5kg
g=10m/s^2

50*.2 + .5*10 = .5*a
a = 30m/s^2

[kzteam5226]

I am still a little confused by the question. The answer is acceleration of 20 m/s^2. However, I am not too sure how this answer was found. Can someone explain the equation of motion which is used for this problem in the answer key? It does not look very familiar.. thanks!

[jsfkt78927]

Because the extra mass is added to the spring at maximum displacement, ie: KE=1/2mv^2 will equal zero.

why, because at maximum displacement the velocity has to be zero.

so remember your consevation of energy equation, E= KE + U

and your spring equations, KE= 1/2mv^2 ; U= 1/2k(delta)x^2.

KE becomes zero and the energy is thus equal to:

(1/2)(50)(4^-2)= 1J

Voila!