Grade 11 Physics Problem About Work and Energy

In summary: I got 8m as well.I thought there was some kind of relation between height and distance. Right now, all I have is Us=1/2 mv^2, which would be 1/2 (25)(x)^2. Since I'm trying to find x, I'd have to find Us first. But I also have a kinetic energy of 800 J. So I'm guessing I'll have to find an equation that relates x and Us first.
  • #1
Felicity26
11
0
1. The energy from elastic potential is 1/2 kx^2. K is a spring constant and x is the displacement distance of the spring from its equilibrium position. How far will a spring of spring constant k=25 N/m compress if an obejct of mass 4 kg has an initial velocity of 20 m/s runs into the spring without slowing down or speeding up?



2. 1/2 kx^2,



3. Basically, all I could come up with was putting the variables into the equation. To be honest, I don't know where to start solving this question or what equations I should use other than the six kinematic equations. The best I can think of after this is converting the mass into Newtons. Actually, what I wanted to know was how do I start solving this question? Sorry:(
 
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  • #2
Felicity26 said:
1. The energy from elastic potential is 1/2 kx^2. K is a spring constant and x is the displacement distance of the spring from its equilibrium position. How far will a spring of spring constant k=25 N/m compress if an obejct of mass 4 kg has an initial velocity of 20 m/s runs into the spring without slowing down or speeding up?



2. 1/2 kx^2,



3. Basically, all I could come up with was putting the variables into the equation. To be honest, I don't know where to start solving this question or what equations I should use other than the six kinematic equations. The best I can think of after this is converting the mass into Newtons. Actually, what I wanted to know was how do I start solving this question? Sorry:(

Conservation of energy. What is the initial energy?
 
  • #3
I would take that 20m/s and find the KE(.5mv^2). then that is converted into "spring energy" solve from there.
 
  • #4
pat666 said:
I would take that 20m/s and find the KE(.5mv^2). then that is converted into "spring energy" solve from there.

Right. Initially the only energy is the kinetic energy of the object. When it compresses the spring to the maximum distance its final speed will be v = 0 (no kinetic energy). So the only energy at the maximum distance compressed is the potential energy in the spring.
 
  • #5
hey, just wondering is the answer 17.9m. also the question is not written well. when you say that it doesn't slow down i assumed that it meant that the final velocity is also 20m/s(y/n) if it doesn't slow down at all the spring will be destroyed and carried away with the projectile.
 
  • #6
pat666 said:
hey, just wondering is the answer 17.9m. also the question is not written well. when you say that it doesn't slow down i assumed that it meant that the final velocity is also 20m/s(y/n) if it doesn't slow down at all the spring will be destroyed and carried away with the projectile.

It says the INITIAL VELOCITY = 20m/s and it doesn't speed up or slow down. All that means is that it stays at v = 20m/s until it hits the spring. No friction is slowing it down and no other force is speeding it up. Once it hits the spring it does start slowing down as the kinetic energy is converted into potential energy of the spring.
 
  • #7
yep that's what i thought, do you have an answer for the question?
 
  • #8
pat666 said:
yep that's what i thought, do you have an answer for the question?

That is not the answer I got. Show your work.
 
  • #9
Sorry, I don't know if the answer is 17.9m; I wasn't given an answer key. I put the numbers in and got 800 J. I was wondering if the equation Eg=mgh would be any help to figuring out the distance?
 
  • #10
Felicity26 said:
Sorry, I don't know if the answer is 17.9m; I wasn't given an answer key. I put the numbers in and got 800 J. I was wondering if the equation Eg=mgh would be any help to figuring out the distance?

Your still not showing your work and thus I cannot help you. Why would gravitational potential energy have to do with this system when it is all on a flat table (presumably)? The height does not change in the problem.
 
  • #11
yeah mgh is static(no movement in the j direction)... 1/2mv^2=1/2kx^2 mv^2=kx^2
4*20^2=25x^2 x=8m not sure how i got 17 or 19 or whatever..
 
  • #12
pat666 said:
yeah mgh is static(no movement in the j direction)... 1/2mv^2=1/2kx^2 mv^2=kx^2
4*20^2=25x^2 x=8m not sure how i got 17 or 19 or whatever..

I got 8m as well.
 
  • #13
I thought there was some kind of relation between height and distance. Right now, all I have is Us=1/2 mv^2, which would be 1/2 (25)(x)^2. Since I'm trying to find x, I'd have to find Us first. But I also have a kinetic energy of 800 J. So I'm guessing I'll have to find an equation that would find Us that includes the kinetic energy? Am I on the right track?
 
  • #14
Felicity26 said:
I thought there was some kind of relation between height and distance. Right now, all I have is Us=1/2 mv^2, which would be 1/2 (25)(x)^2. Since I'm trying to find x, I'd have to find Us first. But I also have a kinetic energy of 800 J. So I'm guessing I'll have to find an equation that would find Us that includes the kinetic energy? Am I on the right track?

No 8m is the answer. Height and distance of what? Do you understand what is happening in this problem? All of the kinetic energy is transferred into elastic potential energy of the spring. At that point the spring is the most compressed and that compression distance is x = 8m. what is Us?
 
  • #15
Sorry, I didn't refresh the page when I submitted the reply. I think I get it now. But, I was wondering, if the equation is 1/2mv^2, why is the half missing? (Oh, I was trying to look up the notation for potential energy and I got U and s as the sub for the spring, but I'm guessing I looked up the wrong thing)
 
  • #16
When I say the missing 1/2, I mean when you put then numbers into solve.
 
  • #17
[tex] E_i = K_i + Ui = K_i + U_gravity + U_spring = (1/2)M(V_i)^2 + Mgh + 0 [/tex]
[tex] E_f = K_f + U_f = K_f + U_gravity + U_spring = (1/2)M(V_f)^2 + Mgh + (1/2)kx^2 [/tex]

[tex]E_i = E_f [/tex]and [tex] V_f = 0 [/tex]

[tex] (1/2)M(V_i)^2 + Mgh = (1/2)kx^2 + Mgh [/tex]

Mgh's cancel out. We could have ignored gravitational potential energy from the beginning since we know it did not change during the problem.

[tex] (1/2)M(V_i)^2 = (1/2)kx^2 [/tex]

Solve for x.
 
  • #18
do the math there is a half on both sides of the equation so you can cancel them, if you prefer leave them it will give you the same answer. hope i was helpful.
 
  • #19
Thank you!
 

1. What is the definition of work in physics?

In physics, work is defined as the amount of force applied to an object multiplied by the distance over which the force is applied. It is a measure of the energy transfer that occurs when a force is exerted and an object is moved in the direction of the force.

2. How is work related to energy?

Work and energy are closely related concepts in physics. Work is the transfer of energy from one object to another, or from one form to another. In other words, work is a way to change an object's energy. The unit of measurement for both work and energy is the joule (J).

3. Can you give an example of a physics problem involving work and energy?

One example of a physics problem involving work and energy is calculating the amount of work done by a force to move an object a certain distance. For instance, if a force of 10 Newtons is applied to move an object 5 meters, the work done would be 50 joules (10N x 5m = 50J).

4. How is the concept of power related to work and energy?

Power is another important concept in physics that is related to work and energy. Power is defined as the rate at which work is done or energy is transferred. It is measured in watts (W) and is calculated by dividing the amount of work done by the time it takes to do the work. In other words, power is a measure of how quickly energy is being transferred or transformed.

5. How can we apply the principles of work and energy in real-life situations?

The principles of work and energy are crucial in understanding many real-life situations. For example, understanding the concept of work and energy is important in designing machines, calculating the efficiency of engines, and explaining the movement of objects. It also helps us understand the concept of conservation of energy, which is the idea that energy cannot be created or destroyed, only transferred or transformed.

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