# Particle movement down a track

1. Oct 25, 2007

### pcandrepair

[SOLVED] Particle movement down a track

1. The problem statement, all variables and given/known data

A particle of mass m = 6.60 kg is released from point A and slides on the frictionless track shown in the figure below. (ha = 5.20 m.)

(a) Determine the particle's speed at points B and C.
point B____________ m/s
point C ___________ m/s

(b) Determine the net work done by the gravitational force in moving the particle from A to C.
____________J

2. Relevant equations

Ef - Ei
mgh + 1/2mv^2 = mgh + 1/2mv^2

3. The attempt at a solution

#### Attached Files:

• ###### p8-24alt.gif
File size:
7.4 KB
Views:
187
2. Oct 25, 2007

### hotcommodity

I'm not seeing your attempt at the solution if you gave one. The particle probably starts from rest, but see if the problem specifies the initial velocity. If it starts at rest, you can eliminate one of the terms in your mechanical energy equation.

Last edited: Oct 25, 2007
3. Oct 25, 2007

### pcandrepair

There is a picture attached but if you can't see it, point B is 3.2m and point C is 2m.

4. Oct 25, 2007

### hotcommodity

I see it now. Can you show me which term gets eliminated in your mechanical energy equation? If you can, you should be able to solve for the final speed of the particle at each point.

5. Oct 25, 2007

### pcandrepair

Since Vo = 0m/s wouldn't the 1/2mv^2 on the left side of the equation cancel and then I would just solve for my Vf on the right side

6. Oct 25, 2007

### SticksandStones

Remember that for energy to be conserved, your potential energy at the start must equal the kinetic energy at the end:
$$mgh = \frac{1}{2}mv^{2}$$

So you have the potential energy from point a to point b and the kinetic energy at point b. How can you use this relationship to find the velocity of the particle?

7. Oct 25, 2007

### hotcommodity

If the particle has an initial speed of zero, then the initial kinetic energy of the particle is zero. You'll only have one variable in the equation to solve for, which is $$v_f$$. Does that make sense?

8. Oct 25, 2007

### pcandrepair

so using mgHa = mgHb + 1/2mVf^2, i got 5.77m/s at point B.

9. Oct 25, 2007

### hotcommodity

I don't have a calculator right in front of me, but it's better that you understand how to get to the right answer than get the right answer. SticksandStones brought up a good point, which is the fact that we're using the conservation of energy here. The only reason you're able to set up the equation as you did was because we ignored friction and air resistance, and therefore mechanical energy was conserved. It looks like you set up the equation correctly, but can you show me what $$v_f$$ is equal to in terms of variables? If you can, you ought to be able to plug in the values you have for the heights, the acceleration due to gravity and the mass, and find the final speed. Which you may have done on paper already.

10. Oct 25, 2007

### pcandrepair

My first answer was wrong cause i solved the formula wrong but i fixed it and found
Vf = sqrt(2(gHa - gHb))

which ended up to be 6.261 m/s.

11. Oct 25, 2007

### hotcommodity

That's the answer I arrived at as well. This is why it's good to set up the equation using variables before we start plugging stuff in. Post again if you have any more questions :)

12. Oct 25, 2007

### pcandrepair

To find part b which is the work done by the gravitational force from point a to c would I use.... mgHa - mgHc = Kinetic energy = Work ?

13. Oct 26, 2007

### hotcommodity

Well, lets be clear about how we're defining work. The work-energy theorem tells us that the net work done on an object is equal to the objects change in kinetic energy, as long as the net force on the object is constant. Remember kinetic energy is defined as .5 times the mass of the object, times its speed squared. So we have:

$$W_{NET} = \Delta KE = .5(m)v^{2}_{f} -.5(m) v^{2}_{i}$$

You've already found the particles speed at C, so we need not deal with the heights any further.