help whit problem of college admission exam

[Camilo08642]

Homework Statement

If we hang an 11 kg object from the ceiling of an elevator using a spring with a spring constant of 650 N/m, what will be the elongation of the spring in centimeters when the elevator is moving upward while decelerating at a rate of 3.3 m/s²

Relevant Equations

Σf=m*a
Fe= Δx*k
w= G*m

Σf=m*a
Fe-w=m*-a is this correct?

 

[jbriggs444]

Σf=m*a
Fe-w=m*-a is this correct?

Certainly, $\sum F = ma$

Next, on the left hand side you are expressing the sum of the forces as the supporting force from the elevator $F_e$ directed upward (positive) and the weight $w$ of the object directed downward (negative).

So far, so good.

On the right hand side you are expressing $m \times a$ as $m \times -a$. That is rather unfortunate notation. You would have us believe that $ma = -ma$.

How about: $F_e - w = ma$ along with $a=-3.3 \ \text{m}/\text{s}^2$

Can you solve for $F_e$?

 

[kuruman]

How about: $F_e - w = ma$ along with $a=-3.3 \ \text{m}/\text{s}^2$

I think that one needs to be careful here with the vectors. Assuming the convention "up" is positive, "down "is negative,

1. Gravity is down: $~\mathbf F_g=mg~(-\mathbf{\hat y})$.
2. We are told that the spring is elongated, so the elastic force is "up": $~\mathbf F_e=F_e~(+\mathbf{\hat y})$.
3. We are told that the elevator is moving up with decreasing speed so its acceleration is down. What about the acceleration of the mass hanging from the spring? If we match its acceleration with that of the elevator and write $~m\mathbf a=ma~(-\mathbf{\hat y})$.

Putting this together gives $$\begin{align} & \mathbf F_e+\mathbf F_g=m\mathbf a \nonumber \\ & F_e~(+\mathbf{\hat y})+mg~(-\mathbf{\hat y})=ma~(-\mathbf{\hat y})\nonumber \\
& F_e-mg=-ma. \nonumber
\end{align}$$ Note that the last equation can be used to find the elongation of the spring when the acceleration of the mass matches that of the elevator.

 

[Camilo08642]

Certainly, $\sum F = ma$

Next, on the left hand side you are expressing the sum of the forces as the supporting force from the elevator $F_e$ directed upward (positive) and the weight $w$ of the object directed downward (negative).

So far, so good.

On the right hand side you are expressing $m \times a$ as $m \times -a$. That is rather unfortunate notation. You would have us believe that $ma = -ma$.

How about: $F_e - w = ma$ along with $a=-3.3 \ \text{m}/\text{s}^2$

Can you solve for $F_e$?

thanks for the answer then the answer would be Δx= (M*(G-a))/k= 11.01 cm

 

[jbriggs444]

thanks for the answer then the answer would be Δx= (M*(G-a))/k= 11.01 cm

The answer seems correct. Though it has too many significant figures. How accurately do you know $M$, $k$, $a$ and local $g$?

It appears that you have used $g$ = 9.80665 m/sec². However, local $g$ will vary significantly from that standard figure.

Gravity on the Earth's surface varies by around 0.7%, from 9.7639 m/s² on the Nevado Huascarán mountain in Peru to 9.8337 m/s² at the surface of the Arctic Ocean.[6] In large cities, it ranges from 9.7806 m/s² [7] in Kuala Lumpur, Mexico City, and Singapore to 9.825 m/s² in Oslo and Helsinki.

 

[kuruman]

I am troubled by the proposed solution of this problem. We are told that

• the spring is elongated
• the elevator is moving up with some undefined velocity $v$
• the acceleration of the elevator, not the mass, is 3.3 m/s² down

So far, so good. What is the justification for equating the acceleration of the elevator with the acceleration of the mass? That would be appropriate if the mass were rigidly attached to the elevator but is not the case here.

Let's consider a plausible scenario that fulfills all of the above pieces of information and see where it leads us. We have an elevator that moves initially at constant velocity $v$ up. The hanging mass is at equilibrium and the elongation of the spring from its relaxed length is $y_0=\dfrac{mg}{k}.~$ At $t=0$ the elevator starts decelerating at 3.3 m/s² as it nears its destination. At that time the mass will start simple harmonic motion. Its displacement from the equilibrium position, subject to the initial conditions, will be $$y(t)=\frac{v}{\omega}\sin(\omega t)~~~~\left(\omega=\sqrt{\frac{k}{m}}\right).$$ Thus, at $t=0$ the elongation of the spring is $$\Delta L=\frac{mg}{k}= \frac{(11\times 9.8)~\text{N}}{650~\text{N/m}}=16.6~\text{cm}.$$ This answer follows from and is consistent with what is given. If this answer is not what the author of the problem intended, then additional information should have been provided.