Max Dist Frame Moves Down from Initial: 15m

[princessfrost]

A 0.150-kg frame, when suspended from a coil spring, stretches the spring 0.050 m. A 0.200-kg lump of putty is dropped from rest onto the frame from a height of 30.0 cm.

Find the maximum distance the frame moves downward from its initial position in meters.




First I found the spring constant, k, which is given by F = -kx.

F = -kx
k = -F/x = -(0.150kg *9.8 m/s2)/(0.050m) = 29.4 N/m

Next, I found the force that the lump of putty makes on the frame. The force is given by F = ma. The mass of the putty, 0.200kg, and the acceleration which is due to gravity, -9.8m/s2.

F = ma = 0.200kg * -9.8m/s2 = 1.96 N

Now the spring constant, k (29.4N/m), and the force of the putty, 1.96N, so I finally solve for the distance the frame moves.

F = -kx
x = -k/F = -(29.4N/m)/(1.96N) = -15m

It is negative, because it moves in the -x direction.

So, the frame moves 15m downwards is what I get but my program says it is wrong. Can someone please help me?

 

[hage567]

This part looks OK:

First I found the spring constant, k, which is given by F = -kx.

F = -kx
k = -F/x = -(0.150kg *9.8 m/s2)/(0.050m) = 29.4 N/m

For the next part, the 1.96 N is not the force that the putty exerts on the frame, since the putty was released 0.3 m above the frame. You can consider conservation of energy to get the distance the spring will go to bring the putty to a stop.

 

[princessfrost]

Ok. so how about:

1/2mv^2= 1/2 kx^2

F= - k x
where F= mg = (0.150-) (9.80m/s2)
F = 1.47 N
1.47 N= -k ( - 0.050 )
k = 1.47N/0.050m
k = 29.4N/m
then:
v22 = v12 - 2gh, where v1 = 0 and h =- 30 cm = - 0.30m
v22 = - 2(9.80m/s2)(- 0.30m)
v22 = 5.88
v2 = 2.425 m/s
Now, plugging these values in I get
mv2 = k x2
x2 = mv2/k = (0.200)(5.88 m2/s2)/(29.4N/m)
x2 = 0.04
x = 0.2 m

So the maximum distance is:
X = 0.05 m + 0.20 m
X = 0.25 m

Is this right?

 

[hage567]

Looks OK to me.