A block is released from rest, determine its velocity

In summary, the problem involves a 1.8-kg block attached to an undeformed spring. When the block is released from rest, its velocity when it hits the floor needs to be determined using the work-energy method. The correct answer is obtained for problem 14.10, but there is confusion with problem 12.47 due to the assumption of ##\int a dx = a \Delta x##. The main topic of chapter 12 is not specified. In solving the problem, the work-energy principle is essentially derived.
  • #1
Alexanddros81
177
4

Homework Statement


14.10 Solve Prob. 12.47 by the work-energy method

12.47 When the 1.8-kg block is in the position shown, the attached spring is
undeformed. If the block is released from rest in this position, determine its velocity
when it hits the floor

Fig P12_47.jpg


Homework Equations

The Attempt at a Solution


Here is my solution to both 12.47 and 14.10.
I get in 14.10 the correct answer but I am confused with 12.47. Is it correct?
Pytels_Dynamics109.jpg
Pytels_Dynamics108.jpg
 

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  • #2
When you wrote
upload_2017-12-18_13-7-12.png

you assumed ##\int a dx = a \Delta x##. Is this true if the acceleration is not constant?

What is the main topic of chapter 12?
 

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  • #3
Pytels_Dynamics111.jpg


Hi! After reading a sample problem of the book I came up with the above solution.
I have a question though: If I integrate both sides of the equation just above equation (3)
shouldn't I be getting on both sides a constant C that cancels out?
 

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  • #4
Alexanddros81 said:
View attachment 217021
I have a question though: If I integrate both sides of the equation just above equation (3)
shouldn't I be getting on both sides a constant C that cancels out?
Each side of the equation would have its own constant of integration. You can combine the two constants into one constant. Then, this constant is determined by the initial conditions as you have done.

Your work here is essentially the derivation of the work-energy principle for this particular problem.
 
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  • #5
Ok thanks for that.
I need to revise my calculus.

Alexandros
 

FAQ: A block is released from rest, determine its velocity

1. What is the initial velocity of the block?

The block is released from rest, meaning its initial velocity is zero.

2. What is the acceleration of the block?

The acceleration of the block can be calculated using the formula a = (vf - vi)/t, where vf is the final velocity, vi is the initial velocity (which is zero), and t is the time. The specific acceleration value will depend on factors such as the block's mass and the force acting on it.

3. How do you determine the velocity of the block at a specific time?

To determine the velocity of the block at a specific time, you can use the formula vf = vi + at, where vf is the final velocity, vi is the initial velocity (which is zero), and a is the acceleration calculated in the previous step. Plug in the values and solve for vf.

4. What is the relationship between time and velocity for a block released from rest?

The relationship between time and velocity for a block released from rest is linear. This means that as time increases, the velocity of the block will also increase at a constant rate, assuming there is no external force acting on the block.

5. How does the surface the block is released on affect its velocity?

The surface the block is released on can affect its velocity by providing friction or resistance. If the surface is rough, it will slow down the block and decrease its velocity. If the surface is smooth, it will allow the block to maintain a higher velocity. Additionally, the angle of the surface can also affect the velocity of the block.

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