# What is the speed of the box at the lower level

1. Feb 14, 2004

### daisy_polly

Hi,

Can anyone help me solve these problems.

The pictures are on the website given on the bottom.

Q1) What is the Ek at the upper level
b) What is the loss in Eg if the box goes from upper level to lower level

c) What is the speed of the box at the lower level.

d) What will the cofficent of friction be if the box stopped in 25m.

Q2) a) What is the total resistance.
b) What is the total current.

http:/lora_polly.tripod.com/

Daisy

2. Feb 14, 2004

### paul11273

What do you have so far? Where do you get stuck?

Do you know the formula for kinetic energy?(check you text.)

What exactly do you mean by Ek and Eg? I assume Ek is kinetic energy of the box. I'm not clear on Eg.

Speed of the box at the lower level? Are we assuming no friction for this one?

How can the box stop in 25m when you show the decline only 5m long? Is this correct?

For Q2, are you missing a couple of resistor values?
What are your formulas for finding parallel and total resistance?(check your text again, then apply both to this problem to find the total resistance).

3. Feb 14, 2004

### daisy_polly

Hi,

By Eg I mean gravitational potential energy and i dont know where to start from for either of the sums for the speed for the speed of an object at lower level there is friction for Q2 i dont think i missed some resistor values.

Thanks
Daisy

4. Feb 14, 2004

### pedestrian

Since you have mass and velocity you can easily solve for the KEi using the equation for it 1/2mV^2 Also since the box goes to the ground it loses all of its potential energy due to gravity(It does not actually lose all of the energy its just converted into KE) so if you solve for its initial PEg then you have your answer.

If your disregarding friction for B then you can solve for it using the Conservation of mechanical energy Mei=Mef. If there is friction you must first find it before you can solve for velocity.

As for C I dont understand how the block can stop in 25m when the incline is only 5m but to solve this you should probably figure out the work done by friction and solve for the force of friction then you can use trig to solve for the angle of the incline and figure out force normal.

Last edited: Feb 14, 2004
5. Feb 15, 2004

### paul11273

Try answering each of your questions one at a time, and make use of your text to find formulas in addition to following Pedestrian's advice in the last post.

For Q2, you must first find the resistance of the parallel portion of the circuit:
Rp=R1*R2/(R1+R2)

Use this for the area of the circuit that has the 9 and 7 (I assume ohms) values.

After you have that value, treat it as one resistor in series with the others. Series resistance is simply the sum of all resistor values. So Rt=R3+R4+R5+Rp

I asked you if you left out any resistor values because I see a total of 5 resistors, but only 3 values. Are we to assume that the two resistors without values next to them are also 6 ohms? How about you check the problem and include the exact text here. That may clear up some questions the Pedestrian and I have asked you about the inconsistancies in the problem.

I would like to see you try these problems and put your work on the board. That way we can help you with your stuck points. You will also learn more if you put time into wresting with it than us just posting a step-by-step procedure for you to follow.

6. Feb 16, 2004

### daisy_polly

Thanks all for your help I have figured how to solve these problems I had a test yesterday and am hoping I will get good marks thanks again for your help.

Daisy