Work done by a Spring Force question

[brittkub1291]

A 220 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.8 N/cm (Fig. 7-30). The block becomes attached to the spring and compresses the spring 11 cm before momentarily stopping.

(a) While the spring is being compressed, what work is done on the block by the gravitational force on it?

(b) What work is done on the block by the spring force while the spring is being compressed?

(c) What is the speed of the block just before it hits the spring? (Assume that friction is negligible.)

(d) If the speed at impact is doubled, what is the maximum compression of the spring?



I know that for the first part W=mg*delta y and the second part should be W=1/2Kx^2 but then i get a little lost on the last two questions. I know this is a spring force and gravitational question but does it also use kinetic? Every time i plug the numbers into the formula for kinetic energy it comes out wrong.



Thanks!

 

[Hootenanny]

Welcome to Physics Forums.

Using the first two equation can you calculate the total work done on the block whilst the spring is being compressed?

And a hint for the next part: Which quantity is conserved whilst the spring is being compressed?

 

[brittkub1291]

So to get the total work i would add the two amounts together.. I'm just a bit confused because it ends up being negative work...

 

[Hootenanny]

So to get the total work i would add the two amounts together.. I'm just a bit confused because it ends up being negative work...

Hmm... The work done by gravity should be positive since the force (weight) is in the same direction as the displacement, the work done by the spring should be negative since the force of the spring is in the opposite direction to the displacement. However, since in this case the work done by gravity is greater than that done by the spring, the total work done should be positive.

Would you mind posting your calculations for the first two parts. Note also that the correct form of the expression for the work done by the spring is -1/2kx² for the reasons I have above.

 

[brittkub1291]

a. W=(.22kg)(9.8m/s^2)(.11m)
W=.23716 J

b.W=-1/2(.028N/m)(.11^2)
and then i get W=-1.694e-4...which confuses me because the correct answer should be -1.694 J...

 

[Hootenanny]

b.W=-1/2(.028N/m)(.11^2)
and then i get W=-1.694e-4...which confuses me because the correct answer should be -1.694 J...

2.8 N/cm \neq 0.028 N/m.

Notice that the units are Newton's divided by centimetres, therefore to convert the constant into N/m you should divide by 0.01m.

Does that make sense?

 

[brittkub1291]

Yeah that does thank you.
I still don't understand how the total work would be positive then, is this correct
.237J-1.694J=-1.457
then would i plug that into the equation for kinetic energy and solve for velocity for part c?

 

[Hootenanny]

Yeah that does thank you.
I still don't understand how the total work would be positive then, is this correct
.237J-1.694J=-1.457
then would i plug that into the equation for kinetic energy and solve for velocity for part c?

Whoops, when I initially calculated the work done by the spring, I thought that the spring constant was in N/m. Your answers are correct.

So, what about the hint I gave in my first post: which quantity is conserved?

 

[brittkub1291]

Well the gravitational energy should have been conserved but when i try .237J=1/2(.22kg)v^2 i get 1.467 m/s, which isn't right.

 

[Hootenanny]

Well the gravitational energy should have been conserved but when i try .237J=1/2(.22kg)v^2 i get 1.467 m/s, which isn't right.

Hmm... you're on the right tracks, but think bigger. HINT: There are no dissipative forces present.

 

[brittkub1291]

Okay the only other force i can think of is that of the earth, so (.22kg)(9.8)=1/2(.22)v^2...which is wrong too.

 

[Hootenanny]

Okay the only other force i can think of is that of the earth, so (.22kg)(9.8)=1/2(.22)v^2...which is wrong too.

I meant that the total energy of the system is conserved. Do you understand why?

 

[brittkub1291]

I think so, but if the total energy is negative how do i find the velocity?

 

[Hootenanny]

I think so, but if the total energy is negative how do i find the velocity?

Can you write out an equation that represents the concept of conservation of energy in this case?

 

[brittkub1291]

Well i think this is what's confusing me. I know that the intial energy plus the work has to equal the final energy, but i don't know which is which.

 

[Hootenanny]

Well i think this is what's confusing me. I know that the intial energy plus the work has to equal the final energy, but i don't know which is which.

Okay, I'll set this up for you:

So, when the spring is fully compressed:
Gravitational PE₁ + Elastic PE₁ + Kinetic Energy₁ = const.

Equally, when the mass leaves the spring:
Gravitational PE₂ + Elastic PE₂ + Kinetic Energy₂ = const.

Subtracting the two equations (and noting that the constants are equal in each case we obtain):

ΔGravitational PE + ΔElastic PE + ΔKinetic Energy = 0

Do you follow?

 

[brittkub1291]

Okay that makes more sense. So Ek=-.237+1.694. Then from there i plug it into the equation Ek=1/2mv^2 and get that v=3.639 m/s. So if the speed is then doubled, i would solve for a new kinetic energy and then using the conservation of energy find the elastic energy, then solve for the compression. Thanks so much!

 

[Hootenanny]

Okay that makes more sense. So Ek=-.237+1.694. Then from there i plug it into the equation Ek=1/2mv^2 and get that v=3.639 m/s. So if the speed is then doubled, i would solve for a new kinetic energy and then using the conservation of energy find the elastic energy, then solve for the compression. Thanks so much!

Sounds good to me