How Does Spring Force Affect Kinetic Energy and Work Done on a Block?

[J-dizzal]

Homework Statement

The block in the figurelies on a horizontal frictionless surface, and the spring constant is 44 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 2.0 N pulls the block in the positive direction of the x axis, stretching the spring until the block stops. When that stopping point is reached, what are (a) the position of the block, (b) the work that has been done on the block by the applied force, and (c) the work that has been done on the block by the spring force? During the block's displacement, what are (d) the block's position when its kinetic energy is maximum and (e) the value of that maximum kinetic energy?

http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/ch0/33.gif
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Homework Equations

F_s=-kx , k=x²/2[/B]

The Attempt at a Solution

i'm trying to first solve for final x postion x_f so that i can find work. which will be equivalent to final kinetic energy.

[/B]

 

[Nathanael]

When the block is at x=2/45 meters then the acceleration of the block will be zero, but the problem asks for where the block stops, which implies the velocity is zero.

 

[SammyS]

Homework Statement

The block in the figurelies on a horizontal frictionless surface, and the spring constant is 44 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 2.0 N pulls the block in the positive direction of the x axis, stretching the spring until the block stops. When that stopping point is reached, what are (a) the position of the block, (b) the work that has been done on the block by the applied force, and (c) the work that has been done on the block by the spring force? During the block's displacement, what are (d) the block's position when its kinetic energy is maximum and (e) the value of that maximum kinetic energy?

http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/ch0/33.gif
[/B]

Homework Equations

F_s=-kx , k=x²/2[/B]

The Attempt at a Solution

i'm trying to first solve for final x postion x_f so that i can find work. which will be equivalent to final kinetic energy.

[/B]

You really should explain your reasoning for the various steps you're taking.

I see you solved an equation to find some value for x. What is significant about this value you found for x.

 

[J-dizzal]

When the block is at x=2/45 meters then the acceleration of the block will be zero, but the problem asks for where the block stops, which implies the velocity is zero.

K=1/2mv^2 where v=0 . but i was not given mass. so i cannot solve for K here.

 

[J-dizzal]

You really should explain your reasoning for the various steps you're taking.

I see you solved an equation to find some value for x. What is significant about this value you found for x.

i erased that equation. i realized that Fs is the spring force not the constant force, so it does not apply to hookes law equation.

 

[Nathanael]

K=1/2mv^2 where v=0 . but i was not given mass. so i cannot solve for K here.

Does the mass make a difference?

 

[J-dizzal]

Does the mass make a difference?

i guess not. so K_f=0 also

 

[J-dizzal]

i guess not. so K_f=0 also

i don't see how to relate the kinetic energies to solving the displacement of the object.

 

[Nathanael]

i don't see how to relate the kinetic energies to solving the displacement of the object.

If the kinetic energy is zero, what happened to all the energy?

 

[J-dizzal]

If the kinetic energy is zero, what happened to all the energy?

its all potential energy?

 

[Nathanael]

its all potential energy?

Talk to me in math

 

[J-dizzal]

Talk to me in math

U=potential engery=work=∫F dx
work=2d for the 2N force
work=∫F_sdx, is F_s=44N/m here? i thought that was the sprink constant k in hookes law?

 

[Nathanael]

U=potential engery=work=∫F dx
work=2d for the 2N force
work=∫F_sdx,

Right, this is the work done by each force. For KE to be zero, they must ____ ?

F_s=44N/m here?

On one side you have units of Newtons and on the other side the units are Newtons/meter, so that's a dead giveaway that the equation can't be correct.

 

[J-dizzal]

Right, this is the work done by each force. For KE to be zero, they must ____ ?
On one side you have units of Newtons and on the other side the units are Newtons/meter, so that's a dead giveaway that the equation can't be correct.

ok, that wouldn't make sense.
so we have two work equations which are simultaneous equations at final x position? it wouldn't make sense to set them equal to each other though.

 

[Nathanael]

ok, that wouldn't make sense.
so we have two work equations which are simultaneous equations at final x position? it wouldn't make sense to set them equal to each other though.

Well in order for the kinetic energy to be zero, they must add to zero (so that way the net work done on the block is zero).

Or you can look at it in different words: (but the idea is the same)
The work done by the external force must equal the change in (KE+PE) of the system. (PE is potential energy from the spring.)

 

[J-dizzal]

Well in order for the kinetic energy to be zero, they must add to zero (so that way the net work done on the block is zero).

Or you can look at it in different words: (but the idea is the same)
The work done by the external force must equal the change in (KE+PE) of the system. (PE is potential energy from the spring.)

i got -1/11 but the correct answer is positive. did i do the math right?

 

[Nathanael]

i got -1/11 but the correct answer is positive. did i do the math right?

It should be positive. Can you show your work?

 

[J-dizzal]

It should be positive. Can you show your work?

2d=-22d²
0=-22d² - 2d
0= 22d² +2d
0= 2d(11d+1)
0=11d+1
-1=11d
d=-1/11

 

[Nathanael]

2d=-22d²

From the equation W_spring+W_{external.force}=0 you get W_spring=-W_{external.force}
However the force of the spring acts in the opposite direction as the displacement, thus it is doing negative work (so your equation is missing an extra negative sign)

Other than this you're good.

 

[J-dizzal]

From the equation W_spring+W_{external.force}=0 you get W_spring=-W_{external.force}
However the force of the spring acts in the opposite direction as the displacement, thus it is doing negative work (so your equation is missing an extra negative sign)

Other than this you're good.

ok thanks.
i don't understand what part d is asking, because the kinetic energy is zero.

 

[J-dizzal]

From the equation W_spring+W_{external.force}=0 you get W_spring=-W_{external.force}
However the force of the spring acts in the opposite direction as the displacement, thus it is doing negative work (so your equation is missing an extra negative sign)

Other than this you're good.

or is it asking for the value of dx here? or would it be 0+dx?

 

[Nathanael]

i don't understand what part d is asking, because the kinetic energy is zero.

The kinetic energy is zero at the end, but on it's way to the end it wasn't zero. The question is asking you what is the maximum value of the kinetic energy and where does this maximum occur.

I have to go for a bit, so I'll leave you to think about how to solve that one.

 

[J-dizzal]

ok thanks again.

 

[SammyS]

ok thanks.
i don't understand what part d is asking, because the kinetic energy is zero.

In post # 17, Nathanael point out the following:

...

The work done by the external force must equal the change in (KE+PE) of the system. (PE is potential energy from the spring.)

In other words: $\displaystyle \ W_\text{external}=(KE+PE) \ .$

Solve for KE & find what x makes it a maximum.

 

[J-dizzal]

solving for KE i got -0.182J.
I don't know why x=0 would not the the x that would make it a maximum because that is when the force of the spring is a minimum. please help thanks.

 

[J-dizzal]

its my understanding that the kinetic energy of the block is at a maximum when its velocity is greatest. In this problem the maximum velocity would happen when the external force is equal to the spring force? x=0 wouldn't be correct because velocity would also be zero. The box would accelerate to a maximum velocity at the distance when the spring force is the same but in the opposite direction.

F_s=F_c
-44N/m(m_x) = 2N
m_x=0.0455m​

 

[Thyphon]

if you pull a motionless spring and then you leave it what happens? ofcourse, harmonic movement...

 

[SammyS]

solving for KE i got -0.182J.
I don't know why x=0 would not the the x that would make it a maximum because that is when the force of the spring is a minimum. please help thanks.

That solution, KE = -0.182J, can only be for some particular value of x. Moreover, if KE is negative, then either v² is negative, or m is negative. Neither of those is achievable in reality.

When I solve, $\displaystyle \ W_\text{external}=(KE+PE) \,,\$ for KE, I get $\displaystyle \ KE=W_\text{external}-PE \ .$

Plug in your expressions for W_external and PE, then find the maximum value.