# Work energy principle and power

• MHB
• Shah 72
In summary, we can calculate the average resistance force experienced by a sky diver with a mass of 80 kg falling from a height of 3000 m and then opening their parachute for the remaining 2000 m of their fall. By using conservation of energy and the equation for work done by an average resistance force, we can find that the average resistance force is approximately 1199 N or 1200 N.
Shah 72
MHB
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

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?

Klaas van Aarsen said:
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.

Klaas van Aarsen said:
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

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?

Last edited:
Klaas van Aarsen said:
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

Klaas van Aarsen said:
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

Klaas van Aarsen said:
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

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 $g=10\text{ m/s}^2$.

Klaas van Aarsen said:
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!

## 1. What is the work energy principle?

The work energy principle states that the work done on an object is equal to the change in its kinetic energy. In other words, the net work done on an object will result in a change in its speed or direction of motion.

## 2. How is work related to energy?

Work and energy are closely related concepts. Work is the transfer of energy from one system to another, and energy is the ability to do work. When work is done on an object, its energy changes.

## 3. What is the formula for calculating work?

The formula for calculating work is W = F * d * cosθ, where W is work, F is the applied force, d is the displacement, and θ is the angle between the force and displacement vectors.

## 4. How is power related to work and energy?

Power is the rate at which work is done or energy is transferred. It is calculated by dividing the work done by the time taken to do it. The unit of power is watts (W), which is equal to one joule per second (J/s).

## 5. How can the work energy principle be applied in real-world situations?

The work energy principle can be applied in various real-world situations, such as calculating the energy needed to lift an object to a certain height, determining the power output of a car engine, or understanding the motion of a pendulum. It is a fundamental concept in physics and is used in many fields, including engineering, mechanics, and thermodynamics.

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