Force and motion -- Person falling from 20m onto a cushion (EASY)

[guest948]

Homework Statement:

A cushion is 3m high when inflated. Assume a person of mass 50kg falls onto the cushion from 20m above the ground and is brought to a stop in 0.25s after hitting the cushion. Assume air resistance is negligible.

How high is the person from the ground when he is stopped by the cushion? Assume that he decelerates uniformly.

TASK: Please help explain why approach 2 is INCORRECT.
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The following have been confirmed:
- The person's velocity just before he hits the cushion is 18.263 ms-1 (downwards).
- The average net force acting on the person during the impact is 3652.62 N (upwards).
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Relevant Equations:

s = (v+u)/2 * t
PE = mgh
KE = mv^2/2
Ft = mv - mu

APPROACH 1 (correct):

Height above ground = 3 - (v+u)/2 * t
= 3 - 18.263/2 * 0.25
= 0.717 m

-----------------------------------------------------
APPROACH 2 (incorrect):

Let d be the height from the ground when he is stopped by the cushion.
PE loss = Work done against motion by cushion
mgh = Fs
mg(20-d) = F(3-d)
50 * 9.81 * (20-d) = 3652.62 * (3-d)
d = 0.363 m

 

[haruspex]

Homework Statement:: A cushion is 3m high when inflated. Assume a person of mass 50kg falls onto the cushion from 20m above the ground and is brought to a stop in 0.25s after hitting the cushion. Assume air resistance is negligible.

How high is the person from the ground when he is stopped by the cushion? Assume that he decelerates uniformly.

TASK: Please help explain why approach 2 is INCORRECT.
----------------------------------------------------------------------
The following have been confirmed:
- The person's velocity just before he hits the cushion is 18.263 ms-1 (downwards).
- The average net force acting on the person during the impact is 3652.62 N (upwards).
----------------------------------------------------------------------
Relevant Equations:: s = (v+u)/2 * t
PE = mgh
KE = mv^2/2
Ft = mv - mu

APPROACH 1 (correct):

Height above ground = 3 - (v+u)/2 * t
= 3 - 18.263/2 * 0.25
= 0.717 m

-----------------------------------------------------
APPROACH 2 (incorrect):

Let d be the height from the ground when he is stopped by the cushion.
PE loss = Work done against motion by cushion
mgh = Fs
mg(20-d) = F(3-d)
50 * 9.81 * (20-d) = 3652.62 * (3-d)
d = 0.363 m

Both methods are wrong since they assume constant acceleration after meeting the cushion. That might be right for a snowpack, but not for an inflated cushion. A spring would be a better model.

In your second method, on the left of the equation you have the work done by gravity over the whole descent. On the right, you have the work done by the net force (which includes gravity) in the last 3-d of the descent. So you are counting the work done by gravity in the last 3-d on both sides.
See what happens if you change the 20-d to 17.

 

[guest948]

.

Both methods are wrong since they assume constant acceleration after meeting the cushion. That might be right for a snowpack, but not for an inflated cushion. A spring would be a better model.

Thanks, but constant acceleration (deceleration) is assumed in this question.

 

[haruspex]

Thanks, but constant acceleration (deceleration) is assumed in this question.

Ok, so see my edit to post #2.