MHB Work energy principle and power

[Shah 72]

A sky diver of mass 80 kg falls 1000m from rest and then opens his parachute for the remaining 2000m of his fall. Air resistance is negligible until the parachute opens. The sky diver is traveling at 5 m/ s just before he hits the ground. Find the average resistance force when the sky diver is falling with the parachute open.
I don't understand this. Pls help

 

[I like Serena]

The work done by an average resistance force $F_{friction}$ is $W_{friction}=F_{friction}\cdot \Delta h$.

Conservation of energy means that $GPE+KE$ just before the parachute opens must be equal to $GPE+KE+W_{friction}$ at the ground.

Can we find those $GPE$'s and $KE$'s?

 

[Shah 72]

The work done by an average resistance force $F_{friction}$ is $W_{friction}=F_{friction}\cdot \Delta h$.

Conservation of energy means that $GPE+KE$ just before the parachute opens must be equal to $GPE+KE+W_{friction}$ at the ground.

Can we find those $GPE$'s and $KE$'s?

Let me try to work it out.

 

[Shah 72]

The work done by an average resistance force $F_{friction}$ is $W_{friction}=F_{friction}\cdot \Delta h$.

Conservation of energy means that $GPE+KE$ just before the parachute opens must be equal to $GPE+KE+W_{friction}$ at the ground.

Can we find those $GPE$'s and $KE$'s?

Before parachute opens
m= 80 kg, h = 1000m, u= 0m/s, v = vm/s
Increase in KE= 1/2mv^2= 40 v^2 J
Loss of GPE= mgh= 80×10× 1000= 800000J
After parachute opens
U= vm/s, v= 5m/s
Increase in KE= 1000- 40v^2
Loss of GPE= 80×10×2000= 1600000J
Work done against resistance = F× 2000 J
I don't know how to calculate after this

 

[I like Serena]

Let's do this a bit more systematically and put it in a table:
\begin{array}{|c|c|c|c|c|c|}
\hline
&h&GPE&KE&\text{Dissipated energy}&\text{Total energy} \\
\hline
\text{Initial} & 3000 & mg\cdot 3000 & 0 & 0 & mg\cdot 3000 \\
\text{Parachute opens} & 2000 & mg\cdot 2000 & mg\cdot 1000 & 0 & mg\cdot 3000 \\
\text{Ground} & 0 & 0 & \frac 12m\cdot 5^2 & F_{friction}\cdot 2000 & \frac 12m\cdot 5^2 + F_{friction}\cdot 2000 \\
\hline
\end{array}
The $\text{Total energy}$ must be the same at all times.
Can we find $F_{friction}$ from that?

 

[Shah 72]

Let's do this a bit more systematically and put it in a table:
\begin{array}{|c|c|c|c|c|c|}
\hline
&h&GPE&KE&\text{Dissipated energy}&\text{Total energy} \\
\hline
\text{Initial} & 3000 & mg\cdot 3000 & 0 & 0 & mg\cdot 3000 \\
\text{Parachute opens} & 2000 & mg\cdot 2000 & mg\cdot 1000 & 0 & mg\cdot 3000 \\
\text{Ground} & 0 & 0 & \frac 12m\cdot 5^2 & F_{friction}\cdot 2000 & \frac 12m\cdot 5^2 + F_{friction}\cdot 2000 \\
\hline
\end{array}
The $\text{Total energy}$ must be the same at all times.
Can we find $F_{friction}$ from that?

I apologize but iam not able to get the ans. Pls help

 

[Shah 72]

Let's do this a bit more systematically and put it in a table:
\begin{array}{|c|c|c|c|c|c|}
\hline
&h&GPE&KE&\text{Dissipated energy}&\text{Total energy} \\
\hline
\text{Initial} & 3000 & mg\cdot 3000 & 0 & 0 & mg\cdot 3000 \\
\text{Parachute opens} & 2000 & mg\cdot 2000 & mg\cdot 1000 & 0 & mg\cdot 3000 \\
\text{Ground} & 0 & 0 & \frac 12m\cdot 5^2 & F_{friction}\cdot 2000 & \frac 12m\cdot 5^2 + F_{friction}\cdot 2000 \\
\hline
\end{array}
The $\text{Total energy}$ must be the same at all times.
Can we find $F_{friction}$ from that?

I calculated using v^2= u^2+2as
To calculate the speed just before the parachute opens.
So I get v= 141.42 m/ s
After the parachute opens
U= 141.42m /s, v= 5m/ s
Increase in KE= -798984.7J
Loss in GPE= mg(h2-h1)= mg (0-2000)= -1600000J
Work done against resistance = -2000F
Increase in mechanical energy = work done
I get 1199 N or 1200N

 

[Shah 72]

Let's do this a bit more systematically and put it in a table:
\begin{array}{|c|c|c|c|c|c|}
\hline
&h&GPE&KE&\text{Dissipated energy}&\text{Total energy} \\
\hline
\text{Initial} & 3000 & mg\cdot 3000 & 0 & 0 & mg\cdot 3000 \\
\text{Parachute opens} & 2000 & mg\cdot 2000 & mg\cdot 1000 & 0 & mg\cdot 3000 \\
\text{Ground} & 0 & 0 & \frac 12m\cdot 5^2 & F_{friction}\cdot 2000 & \frac 12m\cdot 5^2 + F_{friction}\cdot 2000 \\
\hline
\end{array}
The $\text{Total energy}$ must be the same at all times.
Can we find $F_{friction}$ from that?

Pls help me to calculate with the method you have used. Pls

 

[I like Serena]

Please do not bump. That is, please do not post a useless post just to attract attention.

We don't need to calculate the speed.
The energy at the beginning is $mg\cdot 3000$.
We have the same total energy when the parachute opens.
And at the ground the total energy is $\frac 12m\cdot 5^2 + F_{friction}\cdot 2000$.

At each point in time the total energy must be the same. So:
$$mg\cdot 3000 = \frac 12m\cdot 5^2 + F_{friction}\cdot 2000 \implies F_{friction}=\frac{mg\cdot 3000 - \frac 12m\cdot 5^2}{2000}$$
We can now fill in $m=80\text{ kg}$ and $g=10\text{ m/s}^2$.

 

[Shah 72]

Please do not bump. That is, please do not post a useless post just to attract attention.

We don't need to calculate the speed.
The energy at the beginning is $mg\cdot 3000$.
We have the same total energy when the parachute opens.
And at the ground the total energy is $\frac 12m\cdot 5^2 + F_{friction}\cdot 2000$.

At each point in time the total energy must be the same. So:
$$mg\cdot 3000 = \frac 12m\cdot 5^2 + F_{friction}\cdot 2000 \implies F_{friction}=\frac{mg\cdot 3000 - \frac 12m\cdot 5^2}{2000}$$
We can now fill in $m=80\text{ kg}$ and $g=10\text{ m/s}^2$.

Thank you so much!