Frictional force on object falling 1.65 m

[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]

$$\Delta d_{T}$$

$$\Delta d_{P}$$

$$\Delta d_{C}$$

$$\Delta t$$

$${v}$$

$${h}$$

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

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

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

$$m_{t}$$