How High Will Block A Rise After Sudden Platform Stop?

[cheerspens]

Homework Statement

Two blocks are stacked on a platform, which is attached to two cables as shown in the attached figure. If it takes 5.0 sec for the platform to reach a speed of 11.5 m/s directed upward starting from rest find the following information:

After a total of 10 sec of acceleration, the cables suddenly bring the platform to an abrupt stop. Block B is attached to the platform, but block A is free to move. How high above the ground will the block A rise? Not that the platform was initially resting on the ground just before the cables accelerated it upward.

Homework Equations

x=x_o+v_ot+(1/2)at²
v=v_o+at

The Attempt at a Solution

I found the acceleration to be 2.3 m/s².
How should I set this up? I usually start off with a variable list but how do I take into account that block A is going to continue moving?

 

[tomwilliam]

You can do it using energy considerations.
If the kinetic energy of block A when the cable stops is:
(m*v^2)/2 then no when the cables stop no more energy is put into the block, and all the kinetic energy will simply convert to potential energy given by:
mgh.
Does that help?

 

[cheerspens]

I'm not sure that I have enough information to solve it that way using mgh and (1/2)mv^2?

 

[tomwilliam]

Well, there are a few things which complicate this question. Firstly, we don't know whether the acceleration is constant...I guess you'll have to assume it is. That way you can find out the velocity of the system after 10 seconds when the cables stop abruptly. Then, you have the mass of block A, and you know its speed. That means you know the energy of the block at this point. As I say, the energy doesn't change in magnitude after this, it just changes form (into potential energy). So you can equate this magnitude with mgh. You know the mass, and you can take gravitational acceleration to be about 9.81 m s^-2 and therefore calculate h, the height of the block when all its energy is potential (zero kinetic energy, and it has stopped). The extra complication is you have to work out the distance above the ground that the platform has travelled, and add it to the height you've worked out. You don't know the thickness of the platform or the height of the other block, so your answer won't be fully accurate.

 

[AndyK]

Another way you can go with a question like this is the old school kinematic equations. You can cut the problem in half. One half being from the beginning to 5 seconds and the second half being after the 5 second mark. If you calculate the distance between the platform and ground at 5 seconds, then go back and treat block 2 as an object that is thrown upwards with and initial velocity Vo and acceleration 0 and figure out how high it will go and then add the two together and you will have the the total distance between the free block and the ground. This should solve it.
I assume you know how to calculate how high an object will rise before falling back to the Earth if thrown at an angle or directly upwards with initial velocity and 0 acceleration.

 

[AndyK]

You can also solve the energy way as tom said since you have M,V,g, and air resistance is negligible since they did not mention it. But the kinematic way is the first way you learn in the class and a month or so later the energy equation kicks in, so I am not sure if you have covered it yet.

 

[cheerspens]

So would 141.99 meters be correct?

 

[tomwilliam]

Show us your working out.

 

[AndyK]

Yeah show your calculations, and also a clarification to my post, when I said 0 acceleration, actually there is acceleration which is gravity about -9.8m/s^2 if you have a standard coordinate system. I was trying to say that the object is not getting any positive acceleration after the 5 second mark.

 

[cheerspens]

For my answer I got 141.99 meters. Is that correct?

To get this, I solved for the height and velocity the platform would reach after 10 seconds, which was 115 m and 23 m/s. I then took these to solve for how how high the platform would reach (with the X_o being 115 m) and the time it took to get there with the V_o=23m/s, the V=0m/s, and the acceleration -9.8 m/s.
I hope that wasn't a confusing description of how I got to my answer.