# Homework Help: Frictional Force

1. Apr 8, 2007

### raman911

Frictional Fore

m=0.2kg
h1=1.65m
h2=0m
v1=0m/s
v2=4.02m/s

http://img339.imageshack.us/img339/7201/11ei1.png [Broken]

Calculate the frictinal Force

Last edited by a moderator: May 2, 2017
2. Apr 8, 2007

### dvyu

i might be able to help - but what does the 'h' stand for

3. Apr 8, 2007

### Staff: Mentor

Well? What have you done so far?

Hint: There are two ways to go about this: (1) you can calculate the acceleration using kinematics and use that to find the net force, or (2) you can use energy methods.

4. Apr 8, 2007

### raman911

Kinetic Energy1=0J
Kinetic Energy1=2.1J
Gravitational Potential Energy1 =2.3J
Gravitational Potential Energy2 =0J

5. Apr 8, 2007

### raman911

how vcan i find acceleration i haven't time

6. Apr 8, 2007

### dvyu

The net force is equal to mass*acceleration

7. Apr 8, 2007

### raman911

i know but how can i find net force and acceleration

8. Apr 8, 2007

### dvyu

you need to use perpendicular and parallel components

9. Apr 8, 2007

### raman911

Kinetic Energy1=0J
Kinetic Energy1=2.1J
Gravitational Potential Energy1 =2.3J
Gravitational Potential Energy2 =0J

10. Apr 8, 2007

### dvyu

I havent done energy yet at school - I would have to use kinematics, sorry

11. Apr 8, 2007

### raman911

$$Given$$
$$m=0.2kg$$
$$\Delta d=0m$$
$${v}_{1}=0m/s$$
$${v}_{2}=3.96m/s$$
$${h}_{1}=1.65m$$
$${h}_{2}=0m$$

$$Required$$

$${F}_{f}$$

$$Solution$$

$$v_{ave} = (v_1 + v_2)/2$$
$$v_{ave} = (0m/s + 4.02m/s)/2$$
$$v_{ave} = 2.01m/s$$

$$\Delta t = \Delta \vec d/\Delta v_{ave}$$
$$\Delta t = 2.52m/2.01m/s$$
$$\Delta t = 1.25s$$

$$\vec a = \Delta v / \Delta t$$
$$\vec a = (2.01m/s) / 1.25s$$
$$\vec a = 1.608m/s^2$$

$${E}_{T}={m}g\Delta h$$
$$={0.2kg}*9.8N/kg*1.65m$$
$${E}_{g}=3.234J$$

$${E}_{k}=(1/2)m{v}_{ave}^2$$
$$=(1/2)0.2Kg(2.01m/s)^2$$
$${E}_{k}=0.40J$$

$${E}_{T}={E}_{k}+{W}_{f}$$
$$3.234J=0.40J+\vec F_{f}\Delta d$$
$$3.234J=0.40J+\vec F_{f}2.52m$$
$$\vec F_{f}=2.834J/2.52m$$
$$\vec F_{f}=1.112N$$

Last edited: Apr 8, 2007
12. Apr 8, 2007

### raman911

nexxxt?????????/

13. Apr 8, 2007

### Staff: Mentor

You don't need time--you have distance and speed.

14. Apr 8, 2007

### raman911

i have height

i have n't distance

15. Apr 8, 2007

### Staff: Mentor

Hint: If there were no friction, the mechanical energy would be constant. The loss of mechanical energy equals the work done by friction. (I did not check your calculations, but don't round off until the last step. Two digits is not accurate enough.)

16. Apr 8, 2007

### raman911

syou mean distace is 1.65m

17. Apr 8, 2007

### Staff: Mentor

Do you have the angle made by the ramp? (I can't read the writing on the diagram.)

18. Apr 8, 2007

### raman911

i have not angle
only that

m=0.2kg
h1=1.65m
h2=0m
v1=0m/s
v2=4.02m/s

http://img339.imageshack.us/img339/7201/11ei1.png [Broken]

Calculate the frictinal Force

Last edited by a moderator: May 2, 2017
19. Apr 8, 2007

### Staff: Mentor

If you don't have the distance over which the friction acts (or enough info to figure it out) I don't see how you can determine the friction force. (Sorry for not catching that earlier.)

After all, if the ramp were 100 m long you'd need much less friction than if it were 1 m long.

20. Apr 8, 2007

### raman911

d=2.52m.........

21. Apr 8, 2007

### Staff: Mentor

So you do know the distance after all? If so, you'll all set. Find the change in mechanical energy and set it equal to the work done by friction.

22. Apr 8, 2007

### raman911

$${E}_{T}={m}g\Delta h$$
$$={0.2kg}*9.8N/kg*1.65m$$
$${E}_{g}=3.234J$$

$${E}_{k}=(1/2)m{v}_{ave}^2$$
$$=(1/2)0.2Kg(2.01m/s)^2$$
$${E}_{k}=0.40J$$

$${E}_{T}={E}_{k}+{W}_{f}$$
$$3.234J=0.40J+\vec F_{f}\Delta d$$
$$3.234J=0.40J+\vec F_{f}2.52m$$
$$\vec F_{f}=2.834J/2.52m$$
$$\vec F_{f}=1.112N$$
Is that Right Now?

Last edited: Apr 8, 2007
23. Apr 8, 2007

### hage567

Why did you take the average velocity?? v2 is the velocity of the block at the end of the ramp. You need to know the kinetic energy at the end of the ramp.

Why did you start a new post? I see you already have a post for this question. It makes it confusing when there is more than one.

24. Apr 8, 2007

### raman911

so i need to use v2

25. Apr 8, 2007

### hage567

Yes you need to use v2, but in the proper way. Why did you think the average velocity was what you wanted?