Position of Block After Applied Force of 2.8N

In summary, a block attached to a spring with a spring constant of 35 N/m is initially at rest on a frictionless surface at position x = 0. An applied force of 2.8 N pulls the block in the positive x direction, causing the spring to stretch until the block stops. The question asks for the position of the block at the stopping point and suggests using the formula F(spring) = -kx, but the calculated answer of -0.08 is incorrect. The correct answer depends on the direction of the applied force and should be checked using a force diagram. It is also recommended to post questions like this in the homework help section.
  • #1
mochabcha
4
0
I have a problem I need help with:
A block lies on a horizontal frictionless surface and is attached to the free end of the spring, with a spring constant of 35 N/m. Initially, the spring is at its relaxed length and the block is stationary at position x = 0. Then an applied force with a constant magnitude of 2.8 N pulls the block in the positive direction of the x axis, stretching the spring until the block stops.
Assume that the stopping point is reached. What is the position of the block?
 
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  • #2
What's the problem with the problem ? Draw a force diagram. You know how to calculate the force due a stretched spring...do it.
 
  • #3
I've tried the formula F(spring) = -kx and got -.08 but that's not the answer.

It seems to me when the mass stops the applied force will equal the spring force
 
  • #4
next time try posting in the homework help section
 
  • #5
And check your sign. Read what the question says about which direction (positive or negative x-direction) the block is pulled.
 

1. What is the position of the block after an applied force of 2.8N?

The position of the block after an applied force of 2.8N will depend on several factors such as the initial position of the block, the mass of the block, and the surface it is placed on. The force applied can cause the block to move, accelerate, or remain stationary. To accurately determine the position of the block, we need to consider the net force acting on the block and its resulting acceleration.

2. How can we calculate the position of the block after an applied force of 2.8N?

To calculate the position of the block, we can use the equation: x = x0 + v0t + 1/2at2, where x is the final position, x0 is the initial position, v0 is the initial velocity, t is the time, and a is the acceleration. Using this equation, we can calculate the displacement of the block after the applied force and determine its final position.

3. Can the position of the block change if the applied force is increased or decreased?

Yes, the position of the block can change if the applied force is increased or decreased. According to Newton's Second Law of Motion, the acceleration of an object is directly proportional to the net force acting on it. Therefore, if the applied force is increased, the acceleration of the block will also increase, resulting in a larger change in position. Similarly, if the force is decreased, the acceleration and change in position will also decrease.

4. What other factors can affect the position of the block after an applied force?

Apart from the magnitude of the applied force, other factors that can affect the position of the block include the mass of the block, the surface it is placed on, and any external forces acting on it (such as friction or air resistance). These factors can impact the acceleration and ultimately the change in position of the block.

5. Can the position of the block be predicted for different values of applied force?

Yes, the position of the block can be predicted for different values of applied force. By using the equation mentioned in question 2, we can calculate the position of the block for any given applied force, initial position, and other relevant factors. However, it is important to note that the position may vary if there are changes in the external factors affecting the block, as mentioned in question 4.

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