Mass on an Incline with Spring Attached

[whoareyou]

Homework Statement

Homework Equations

Forces and/or energy

The Attempt at a Solution

I honestly don't know where to start. First I tried kx = mgsin(tan-1 (3/4))) and got the answer E, which is wrong. Then I tried using energy, mgh = 1/2kx^2 which resulted in an answer that wasn't on the list. Any ideas on how to begin?

 

[rude man]

At any point after letting go of the mass,
energy stored in spring + potential energy + kinetic energy = constant. (Why?)

 

[whoareyou]

Because total mechanical energy of the system is conserved?

Edit: nvm I figured it out!

 

[CAF123]

Why is $kx = mg \sin \theta$ wrong here?, $\theta$ the inclination of ramp.

 

[rude man]

Why is $kx = mg \sin \theta$ wrong here?, $\theta$ the inclination of ramp.

Well, it violates the law of conservation of energy.

If you slowly let the mass slip until it was stationary then you'd be right, but that is not what's happening here.

 

[CAF123]

Well, it violates the law of conservation of energy.

If you slowly let the mass slip until it was stationary then you'd be right, but that is not what's happening here.

Perhaps I misinterpreted something: when the mass is released, it obeys $mg\sin\theta - kx = ma_x$So if they were ever equal, then the block would then move at constant speed.

The energy approach would indeed seem more appropriate.

 

[rude man]

Perhaps I misinterpreted something: when the mass is released, it obeys $mg\sin\theta - kx = ma_x$So if they were ever equal, then the block would then move at constant speed.

The energy approach would indeed seem more appropriate.

Indeed it is.

The point of where the mass momentarily stops moving is not the point of zero acceleration. It is the point of zero velocity.

 

[CAF123]

Yes, but what does the acceleration vectors look like in such a case? It can't be like the vectors on a pendulum, so the acc. vectors will lie along direction of motion.

 

[rude man]

Yes, but what does the acceleration vectors look like in such a case? It can't be like the vectors on a pendulum, so the acc. vectors will lie along direction of motion.

Let s = distance along ramp starting from the point where the spring is relaxed, where s = 0:
Then form F = ma:
-ks + mg sin(θ) = md²s/dt²
Solve this differential equation for s, take ds/dt and then d²s/dt² gives you your acceleration. You will find s(t) and its derivatives are non-decaying oscillations.

You can also solve the given problem this way instead of invoking energy arguments: solve for ds/dt and set ds/dt = second zero (the first is obviously when you let go of the mass). Then solve for T, the time of the second zero of ds/dt, and then s(T).

 

[Tee01]

Perhaps I misinterpreted something: when the mass is released, it obeys $mg\sin\theta - kx = ma_x$So if they were ever equal, then the block would then move at constant speed.

The energy approach would indeed seem more appropriate.

how would you find ma_x?

 

[Delta2]

how would you find ma_x?

You can't find it directly by other means, other than solving the differential equation, that is you set by definition $a_x=\frac{d^2x}{dt^2}$ and solve the differential equation $$mg\sin\theta-kx=m\frac{d^2x}{dt^2}$$.

 

[Delta2]

However this problem can be solved by energy method without solving a differential equation, just an algebraic equation.
$$mgh=\frac{1}{2}kx^2$$
$$h=x\sin\theta$$
Since this is quite an old thread, i suppose i can give the answer, it is C. 0.235m

 

[Tee01]

However this problem can be solved by energy method without solving a differential equation, just an algebraic equation.
$$mgh=\frac{1}{2}kx^2$$
$$h=x\sin\theta$$
Since this is quite an old thread, i suppose i can give the answer, it is C. 0.235m

So for this one, would you use the same method? I'm somehow getting 0.1178 for 2b and 0.245 for 2a.

 

[Delta2]

I did it and i get 0.5m for a) and 0.24m for b). used g=10m/s^2. Please post your workings here if you want me to comment on them.

 

[Tee01]

My friend did 2a and i did 2b. Both are far off. First image is 2a and second image is 2b

 

[Delta2]

I thought we were going to do it by energy method but you are using force method and by doing so you are wrong. You calculate the position where the total force is zero, hence where the acceleration is zero. But we want the position where the velocity is zero.

 

[Tee01]

I thought we were going to do it by energy method but you are using force method and by doing so you are wrong. You calculate the position where the total force is zero, hence where the acceleration is zero. But we want the position where the velocity is zero.

Do you equal potential energy to kinetic energy?

 

[Delta2]

nope. For a) i equal gravitational potential energy to elastic potential energy of the spring.
for b) i use work energy theorem $$W_{TOTAL}=\Delta K$$. We know that $$\Delta K=0-0$$ since the block starts from rest, and the end position is also one with zero speed. So calculate the total work and set it to zero. You ll have one equation with one unknown, the displacement of the spring x.

 

[Delta2]

Better use work energy theorem for a) too . The logic is the same as for b), it is just that the total work is different in each case.

 

[Tee01]

nope. For a) i equal gravitational potential energy to elastic potential energy of the spring.
for b) i use work energy theorem $$W_{TOTAL}=\Delta K$$. We know that $$\Delta K=0-0$$ since the block starts from rest, and the end position is also one with zero speed. So calculate the total work and set it to zero. You ll have one equation with one unknown, the displacement of the spring x.

I'm a little confused :(. I don't know what W_total is.

 

[Delta2]

I'm a little confused :(. I don't know what W_total is.

It is the total work done on the block. Haven't you been introduced to the concept of work , and to work-energy theorem yet?

 

[Tee01]

It is the total work done on the block. Haven't you been introduced to the concept of work , and to work-energy theorem yet?

Also, did you use energy theorem for the original post?

 

[Tee01]

I got 0.0589 on 2b when i do xsin30. My answer is /10 by the right answer and i also got 0.0235 instead of 0.235 as the answer to the original post

 

[Delta2]

No i used conservation of energy actually for the OP, but it can be done with work-energy theorem as well, we ll get almost the same equations.

 

[Tee01]

Oh i confused my 2b with my 2a

 

[Delta2]

I got 0.0589 on 2b when i do xsin30. My answer is /10 by the right answer and i also got 0.0235 instead of 0.235 as the answer to the original post

First of all tell me the answer key for a) and b) please.
Second what do you mean by "do xsin30".
And show me your working for the OP.

 

[Tee01]

However this problem can be solved by energy method without solving a differential equation, just an algebraic equation.
$$mgh=\frac{1}{2}kx^2$$
$$h=x\sin\theta$$
Since this is quite an old thread, i suppose i can give the answer, it is C. 0.235m

x represents the extension right?

 

[Delta2]

yes.

 

[Tee01]

I'm a little confused :(. I don't know what W_total is.

Does Delta K represent the change in spring constant?

 

[Delta2]

No $\Delta K$ is the change in kinetic energy of the block. I use the capital letter $K$ for kinetic energy. The one for the spring constant is the small letter $k$.