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

[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}