Calculating Frictional Force: m=0.2kg h1=1.65m h2=0m v1=0m/s v2=4.02m/s"

[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

Calculate the frictinal Force

 

[dvyu]

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

 

[Doc Al]

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.

 

[raman911]

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

 

[raman911]

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.

how vcan i find acceleration i haven't time

 

[dvyu]

The net force is equal to mass*acceleration

 

[raman911]

The net force is equal to mass*acceleration

i know but how can i find net force and acceleration

 

[dvyu]

you need to use perpendicular and parallel components

 

[raman911]

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

 

[dvyu]

I haven't done energy yet at school - I would have to use kinematics, sorry

 

[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

 

[raman911]

nexxxt?/

 

[Doc Al]

how vcan i find acceleration i haven't time

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

 

[raman911]

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

i have height

i have n't distance

 

[Doc Al]

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

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.)

 

[raman911]

syou mean distace is 1.65m

 

[Doc Al]

i have height

i have n't distance

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

 

[raman911]

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

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

Calculate the frictinal Force

 

[Doc Al]

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.

 

[raman911]

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.

d=2.52m...

 

[Doc Al]

d=2.52m...

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.

 

[raman911]

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.

{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?

 

[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.

 

[raman911]

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.

so i need to use v2

 

[hage567]

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

 

[Doc Al]

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

This looks good. Initially, the energy of the mass is purely PE (KE = 0).

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

Why did you use the average velocity?? You need the final energy when the mass reaches the bottom of the ramp (PE = 0).

{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?

The method is correct, but the answer is not--redo the final KE calculation.

(Note: Threads merged--one thread per problem, please!)​

 

[raman911]

This looks good. Initially, the energy of the mass is purely PE (KE = 0).


Why did you use the average velocity?? You need the final energy when the mass reaches the bottom of the ramp (PE = 0).


The method is correct, but the answer is not--redo the final KE calculation.

(Note: Threads merged--one thread per problem, please!)​

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


{E}_{k}=(1/2)m{v}^2
=(1/2)0.2Kg(4.02m/s)^2
{E}_{k}=1.61J

{E}_{T}={E}_{k}+{W}_{f}
3.234J=1.61J+\vec F_{f}\Delta d
3.234J=1.61J+\vec F_{f}2.52m
\vec F_{f}=1.61J/2.52m
\vec F_{f}=0.64N
Is that Right Now?

 

[Doc Al]

Looks good to me.

 

[raman911]

Energy

If No mechanical Energy Was Lost How Would The Gravitational Energy At The Top Compare With The Kinetic Energy At The Bottom?

 

[Doc Al]

What do you think?

 

[raman911]

. In height one Gravitational energy is equal Total energy because object at top. In height two Kinetic energy is equal Total energy because object at the bottom.

 

[Doc Al]

OK, but how does that answer the question?

 

[raman911]

OK, but how does that answer the question?

can u give me some hints.

 

[Doc Al]

I gave you a big hint in post #15.

 

[raman911]

I gave you a big hint in post #15.

i got it...
thaxxxxxxxx

 

[raman911]

{Fe}_{(s)}+{CuCl}_{2}-{FeCl}_{2}+{Cu}_{(s)}

M_{fe}=55.85g/mol
m=1g
n = \frac{m}{M}
n = \frac{1g}{55.85g/mol}
n = 0.0179mol

\frac{1~mol~of~Fe}{0.0179~mol~of~Fe}=\frac{1~mol~of Cu}{X~mol~of~ Cu}

x = 0.0179mol

M_{cu}=63.55g/mol

m=n*m

m=0.0179mol*63.55g/mol

m=1.14g

Mass~of~Copper=1.14g

Theoretical~mass=1.14g

Actual~mass=1.08g

yield=\frac{Actual~mass}{Theoretical~mass}{*} 100}

 

[raman911]

Given
m=0.2kg
m_{t}=0.750kg
{h}=0.86m
{v}=0.97m/s
~Powered~ Phase~ \Delta d=4.53m
~Coasting~ Phase~ \Delta d=2.02m


Required

{E}_{k},Power,{F}_{f} ~On ~Coasting ~Phase ~and~{F}_{f} ~On ~Powered ~Phase

Solution

Initial Energy= Gravitational Potential Energy
{E}_{g}={m}g\Delta h
={0.2kg}*9.8N/kg*0.86m
{E}_{g}=1.68J

Maximum Kinetic Energy
{E}_{k}=(1/2)m{v}^2
=(1/2)0.200Kg(0.97m/s)^2
{E}_{k}=0.1J

Calculating ~{F}_{f} ~On ~Powered ~Phase

\vec a=v/ \Delta t
\vec a=\frac{0.97m/s}{4.63s}}
\vec a=0.21m/s^2

\vec F_{app}=mg
\vec F_{app}=0.200Kg*9.8N/Kg
\vec F_{app}=1.96N

\vec F_{Net}=m \vec a
=0.750Kg*0.21m/s^2
\vec F_{Net}=0.16N

\vec F_{Net}=\vec F_{app}+\vec F_{f}
\vec F_{f}=0.16N-1.96N
\vec F_{Net}=-1.80N
\vec F_{Net}=1.80N


Calculating ~{F}_{f} ~On ~Coasting ~Phase

{E}_{T}={m}g\Delta h
={0.2kg}*9.8N/kg*0m
{E}_{T}=0J

{E}_{T}={E}_{k}+{W}_{f}
0J=0.35J+\vec F_{f}\Delta d
0J=0.35J+\vec F_{f}*2.02m
\vec F_{f}=-0.35J/2.02m
\vec F_{f}=-0.17N

 

[raman911]

Average~Maximum ~of~Speed~ Car=\frac{1.21m/s+0.90m/s+0.81m/s}{3}
Average~Maximum~of ~Speed~ Car=0.97m/s

Average~Time~ During~ Powered ~Phase =\frac{4.94m+4.35m+4.59m}{3}
Average~Time~ During~ Powered ~Phase =4.63m

Average~Time~ During~ Coasting ~Phase =\frac{1m+3.45m+1.60m}{3}
Average~Time~ During~ Coasting ~Phase =2.02m

Average~Total~Distance=\frac{7m+7.35m+5.30m}{3}
Average~Total~Distance=6.55m

 

[raman911]

\Delta d_{T}

\Delta d_{P}

\Delta d_{C}

\Delta t

{v}

{h}

\Delta d_{C}=\Delta d_{T}-\Delta d_{P}

{v}=\Delta d_{P}/\Delta t

{v}~and~\Delta d_{C}

m_{t}