How to find the distance attained by a sphere after being compressed by a spring?

Jun 2017
337
6
Lima, Peru
The problem is as follows:

A smooth sphere with negligible friction of $m=100\,g$ is released from $A$. The spring compresses to $10\,cm$. As a result the sphere reaches $B$, but at this point it is measured the velocity to be zero. Given this conditions. Find the distance in meters between $A$ and $B$. Assume: $K=600\,\frac{N}{m}$ and $g=10\,\frac{m}{s^2}$.



The alternatives are as follows:

$\begin{array}{ll}
1.&\textrm{3 m}\\
2.&\textrm{4 m}\\
3.&\textrm{5 m}\\
4.&\textrm{6 m}\\
4.&\textrm{10 m}\\
\end{array}$

What I've attempted to do was to establish the conservation of mechanical energy as follows:

$E_p=E_g$

$mgh=\frac{1}{2}kx^2$

Now the problem arises from which length should be used in consideration here?.

What I did was:

$(0.1)(10)(x\sin 37^{\circ})=\frac{1}{2}(600)\left(10\times10^-2\right)^2$

Simplifying this expression I obtained:

$x=5\,m$.

Which corresponds to the answer. But I'm not very convinced of my method. Does it exist another way to do this problem?. In this case the height which I'm assuming if counted from point $A$ to point $B$ and not exactly from where the spring is compressed. But does it exist a justification for this?.

Why the compression of the spring is not expressed in the $y$ direction as $10\cos 37^{\circ}$?. I've already attempted that but it did not work.

Can someone help me here please?.
 
Last edited:

topsquark

Math Team
May 2013
2,443
1,012
The Astral plane
The problem is as follows:

A smooth sphere with negligible friction of $m=100\,g$ is released from $A$. The spring compresses to $10\,cm$. As a result the sphere reaches $B$, but at this point it is measured the velocity to be zero. Given this conditions. Find the distance in meters between $A$ and $B$. Assume: $K=600\,\frac{N}{m}$ and $g=10\,\frac{m}{s^2}$.



The alternatives are as follows:

$\begin{array}{ll}
1.&\textrm{3 m}\\
2.&\textrm{4 m}\\
3.&\textrm{5 m}\\
4.&\textrm{6 m}\\
4.&\textrm{10 m}\\
\end{array}$

What I've attempted to do was to establish the conservation of mechanical energy as follows:

$E_p=E_g$

$mgh=\frac{1}{2}kx^2$

Now the problem arises from which length should be used in consideration here?.

What I did was:

$(0.1)(10)(x\sin 37^{\circ})=\frac{1}{2}(600)\left(10\times10^-2\right)^2$

Simplifying this expression I obtained:

$x=5\,m$.

Which corresponds to the answer. But I'm not very convinced of my method. Does it exist another way to do this problem?. In this case the height which I'm assuming if counted from point $A$ to point $B$ and not exactly from where the spring is compressed. But does it exist a justification for this?.

Why the compression of the spring is not expressed in the $y$ direction as $10\cos 37^{\circ}$?. I've already attempted that but it did not work.

Can someone help me here please?.
I didn't check the numbers but the method is correct. (Though I would have explicitly made a statement that you were setting a zero level at A.) If you really want to you can track the speed of the ball by using a method using a free body diagram, but when you get done with it you will have the same equation anyway, so nothing is gained by this.

As to the spring, the direction of the force is always in line with the path the particle has taken. Thus it will be in line with the incline, meaning that the angle of the incline does not have an effect on the spring part of the energy.

-Dan
 

skeeter

Math Team
Jul 2011
3,266
1,766
Texas
The problem clearly states the ball is "released" at point A. If the spring were not already compressed at that position, it wouldn't move toward B.