An airplane pilot fell 380 m after jumping from an aircraft without his parachute opening. He landed in a snowbank, creating a crater 1.4 m deep, but survived with only minor injuries. Assuming the pilot's mass was 80 kg and his terminal velocity was 50 m/s, estimate the following.
(a) the work done by the snow in bringing him to the rest
(b) the average force exerted on him by the snow to stop him
(c) the work done on him by the air resistance as he fell
thanks. i got A and C. im trying to figure out B
thank you again!
Since Work(W) is Force(F) times Distance(D), (W=F*D), if you know W and D, you should be able to figure F.
To estimate the requested values, we need to consider the energy changes and forces involved in the situation.
(a) The work done by the snow in bringing him to rest:
When the pilot falls and lands on the snowbank, the snow exerts a force to stop his fall. Work is defined as the product of force and the distance over which the force is applied. In this case, the force acting on the pilot is the force exerted by the snow, and the distance is the depth of the crater created by the impact.
The work done by the snow can be calculated using the formula: Work = Force x Distance.
Given:
Force = ?
Distance = 1.4 m
We need to determine the force exerted by the snow on the pilot. To do that, we can use the concept of force being related to the change in momentum. As the pilot falls and is brought to rest, his momentum changes from an initial value of m x v (mass x initial velocity) to zero. Using the principle of conservation of momentum and assuming no other external forces act on the pilot, we can equate the initial momentum to the final momentum:
Initial momentum = Final momentum
m x v = 0
Therefore, Force x time = m x Δv
Force = (m x Δv) / time
The change in velocity (Δv) can be determined by subtracting the final velocity (which is zero) from the initial velocity of the pilot. The time it takes for the pilot to come to rest can be calculated using the formula Distance = (initial velocity x time) + (1/2 x acceleration x time^2), where the acceleration is the acceleration due to gravity. Rearranging the equation, we can solve for time.
We are given:
Delta v = -v (taking the upward direction as positive)
Mass (m) = 80 kg
Initial velocity (v) = 50 m/s
Distance (d) = 1.4 m
Using the equation for distance, we can substitute the given values:
1.4 = (50 x t) + (1/2 x 9.8 x t^2)
Solving this quadratic equation, we find that t ≈ 0.2132 seconds.
Now, we can calculate the force exerted by the snow:
Force = (m x Δv) / time
Force = (80 kg x -50 m/s) / 0.2132 s
Force ≈ -18,864 N
Here, the negative sign indicates that the force exerted by the snow is opposite to the downward direction.
Finally, substituting the values into the work formula:
Work = Force x Distance
Work = -18,864 N x 1.4 m
Work ≈ -26,390 J
The work done by the snow to bring the pilot to rest is approximately -26,390 Joules.
(b) The average force exerted on him by the snow to stop him:
The average force exerted on the pilot is the force required to bring him to rest over the distance traveled. It can be calculated using the formula: Average Force = Work / Distance.
Given:
Work = -26,390 J
Distance = 1.4 m
Average Force = -26,390 J / 1.4 m
Average Force ≈ -18,850 N
The average force exerted on the pilot by the snow to stop him is approximately -18,850 Newtons.
(c) The work done on him by the air resistance as he falls:
When the pilot falls, air resistance opposes his motion. The work done by air resistance can be calculated as the product of the force of air resistance and the distance traveled.
Since the pilot fell without a parachute, we can assume he is in freefall and his velocity approaches a constant terminal velocity. At terminal velocity, the upward force of air resistance is equal to the downward force of gravity, resulting in a net force of zero. Therefore, the work done by air resistance is zero since there is no displacement in the direction of the force.
Therefore, the work done on the pilot by air resistance as he fell is zero.
work by snow: has to equal 1/2 m vterminal^2
average force: Force*distance=work done
average work done on him? Only God knows what that means. averagework/second? avgwork/meter?
work done=changein Potential energy-finalKE
I have no idea what average means here.