# Work problem 2

1. Mar 13, 2010

### cyspope

1. The problem statement, all variables and given/known data

71. A paratrooper fell 370 m after jumping from an aircraft without his parachute opening. He landed in a snowbank, creating a crater 1.1 m deep. Assuming the paratrooper's mass was 80 kg and his terminal velocity was 30 m/s, estimate : a) the work done by the snow in bringing him to rest; b) the average force exerted on him by the snow to stop him; c) the work done on him by air resistance as he fell.

A. a) -36000 J , b) -3300 N , c) -250000 J

2. Relevant equations
W = mgh = Fdsin$$\theta$$= $$\frac{1}{2}$$mv$$^{2}$$

3. The attempt at a solution
a) I got a) by using .5mv$$^{2}$$. However, I don't know how to get b) and c)

2. Mar 13, 2010

### godtripp

what's making the paratrooper fall? Force by gravity. So to stop him what force must be applied? Think of newtons third law.

What is work? work is force times distance. The distance is the depth of the crater, what is the work?

3. Mar 14, 2010

### PhanthomJay

Where are these answers coming from? None are correct........

4. Mar 14, 2010

### cyspope

newton's third law is about action and reaction, isn't it?

this is what i did for b)
W = Fdsin$$\theta$$ = 80 * 9.8 * 1.1 = 862.4 Joule

However, this is wrong because my answer is different from the given answers.

5. Mar 14, 2010

### PhanthomJay

your answer and the given answers are incorrect. The net work done on the 'trooper in bringing him to rest is his change in KE. The net work includes the work done by the snow and the work done by gravity. Once you calculate the work fone by the snow, then the force of the snow on the trooper can be calculated using the definition of work. Are you familiar with conservation of energy equations? Otherwise, you'll have to use the kinematic equations and newtons laws.

6. Mar 14, 2010

### godtripp

What is work?

$$Work=\int F ds$$

From this you can find that

$$Work=-\Delta U$$ and $$Work=\Delta KE$$