# Thread: Using energy to find velocity

1. ## Using energy to find velocity

An inclined slope at 30deg is smooth and A has mass 5kg whilst B has mass 20kg.

If the system is released from rest, use an energy balance equation to find the speed with which B hits the ground.

A is on the slope
B is hanging free.

where do i start?

2. What is the height of B from the ground?

You have to find the acceleration a of the system.

Then fall of potential energy = rise in kinetic energy.

3. 2m from ground

4. What are the forces acting on 20 kg and 5 kg?

Apply Newton's law and find the acceleration of the masses. Magnitude of the accelerarion for each mass is the same.

5. but that's not using energy to v right?
i know how to use newton's law to find a but are you sure that's using energy.

A - T-5gsin30=5a
B - 20g-T=20a

T=58.8N
a=6.86ms^-2

v^2=2as
v=5.24ms^-1

what's the energy way im bit confused.

thanks for the help!

6. A couple of questions: are the masses A and B connected with, say, a rope? I'm guessing that's so based on the phrase "B is hanging". How long is the rope? Is it obvious that A will remain on the slope? Or will the mass B pull A over the top?

7. does this help?
that's all im given

8. Fall in PE = rise in KE

m*a*s = 1/2*m*v^2.

9. Yeah, I think we can assume that B will not pull A over the edge.

So, to use the energy method, you must write down the conservation of energy equation. In order to do that, you have to define events. We'll call releasing the system event 1, with the corresponding subscripts. We'll call B hitting the ground event 2, with corresponding subscripts. Using $T$ for kinetic energy and $U$ for potential energy, we have the following equation:

$T_{A1}+T_{B1}+U_{A1}+U_{B1}=T_{A2}+T_{B2}+U_{A2}+U _{B2}.$

Here the additional subscripts for $A$ and $B$ should be obvious.

Now, right off the bat, you know that $T_{A1}=T_{B1}=0,$ because the system is released from rest. That gives you the equation

$U_{A1}+U_{B1}=T_{A2}+T_{B2}+U_{A2}+U_{B2}.$

Furthermore, since A and B are connected by a rope that is presumably not stretchy, then A and B will be going at the same velocity when B hits the ground. That does not give us equality in the kinetic energy, though, since the masses are different. It is something to remember, though. That is, $v_{A2}=v_{B2}$. Let's call the velocity $v$. So the equation becomes

$U_{A1}+U_{B1}=\dfrac{1}{2}\,(m_{A}+m_{B})\,v^{2}+U _{A2}+U_{B2}.$

Now, in order to find the potential energy terms, you must define a coordinate system. What do you want to use?

10. oh i understand now.

using this

$U_{A1}+U_{B1}=\dfrac{1}{2}\,(m_{A}+m_{B})\,v^{2}+U _{A2}+U_{B2}.$

$0+20*9.8*2=\dfrac{1}{2}\,(5+20)\,v^{2}+5*9.8*1+0.$

thanks really helped me to understand.

11. Since it is the change in potential energy that is physically meaningful, and not its absolute value anywhere, your equation is correct.

You're welcome. Have a good one!