Max Dist Frame Moves Down from Initial: 15m

• princessfrost
Putting it all together:In summary, 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. Using the spring constant formula, the spring constant is found to be 29.4 N/m. The force exerted by the putty on the frame is calculated to be 1.47 N using the formula F = mg. Using conservation of energy, the maximum distance the frame moves downward is found to be 0.25 m.
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?

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.

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?

Looks OK to me.

1. What does "Max Dist Frame Moves Down from Initial: 15m" mean?

This statement refers to the maximum distance that a frame has moved downwards from its initial position, which is 15 meters.

2. What is a "frame" in this context?

In this context, a frame is a structural element that is used to support or enclose something, such as a building or a device. It can also refer to a specific unit of measurement for distance.

3. How is the maximum distance calculated?

The maximum distance is typically calculated by measuring the vertical distance between the initial position of the frame and its lowest point.

4. What does this measurement tell us?

This measurement can provide important information about the stability and structural integrity of the frame. It can also help engineers and builders determine the appropriate design and materials for construction.

5. Why is this measurement important?

The maximum distance that a frame can move downwards from its initial position is an important factor in ensuring the safety and durability of a structure. It can also help determine the potential impact of external forces, such as earthquakes or strong winds, on the frame.

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