# Homework Help: Weird physics question

1. May 23, 2010

### physics(L)10

1. The problem statement, all variables and given/known data
A block, moving along a horizontal plank, is acted upon by a force of friction that is equal to 35% of the weight of the block. The block comes to rest from a speed of x m/s. Find x.

2. Relevant equations
v=u + at
F=F1 + F2

3. The attempt at a solution

I can't even attempt this because I have no clue where to start. They give you basically no information

Last edited: May 23, 2010
2. May 23, 2010

### pgardn

They give you information in the form of variables. They want an answer in the form of variables. Just to make sure, are they using x to represent speed, or x to represent distance traveled horizontally? This makes a difference.

3. May 23, 2010

### physics(L)10

They are using x to represent speed

Attempt (most likely wrong):

Equation to use: F=ma

You can find a relative mass (35% of the block weight) that stops the block. So that will =m. Force I don't know and a I don't know. I'm guessing you find a relative value of a which then can be used to find the speed.

4. May 23, 2010

### pgardn

Well you have two forces on the mass in the vertical, do you know what these are? They must be equal in magnitude and opposite in direction since the mass is not accelerating in the vertical.
And one force acting in the horizontal, friction. They are telling you that the frictional force is a portion of one of the forces in the vertical.

Here would be a picture of the forces assuming the mass is moving to the right. Do you know what the different forces are called. If so we can figure out a and get this done with some easy kinematics.

5. May 23, 2010

### physics(L)10

The down force is the block pushing against the floor and the up force is the ground pushing back against the block (normal force), correct?

6. May 23, 2010

### pgardn

Almost, the down force is due to gravity, mg... the earth pulling on the mass... and yes, the ground supplies the normal force back on the block. So now what can we say about the frictional force in terms of mg?

7. May 23, 2010

### physics(L)10

35%mg?

8. May 23, 2010

### pgardn

yep or 0.35mg which is equal to the frictioanl force which you already called ma. So now solve for a...

9. May 23, 2010

### physics(L)10

v=u + at

where v=final velocity=0, u=initial velocity, a=acceleration=.35mg, t=time=20 seconds

I forgot to add the time into the original question. My bad.

10. May 23, 2010

### pgardn

hold on...

0.35mg = ma so the m's cancel right? so after this you got a, v, t, and you are trying to determine u, which for some reason I cant figure is being called x? So you got everything you need.

11. May 23, 2010

### physics(L)10

So (.35)(9.8)=a, v=0, t=20 and then we get u :)

12. May 23, 2010

### pgardn

yep...

so apparently friction is causing your mass to accelerate in the opposite direction it is moving bringing it to a halt v =0 over a 20 sec time period.

13. May 23, 2010

### physics(L)10

Yuppp I get it :d thank sooooo much