Work, Energy and Momentum

[LBT]

Here is the first question that I can't solve...

A guy, whose mass is 80 kg, jumps off a bench that is 1 meter off the ground. Immediately after landing on the ground with both feet, he jumps 1 meter up into the air. What is the average force on his left foot if the time of contact with the ground is 0.2 seconds?

anyone got an idea???

thanks

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[prophecygirl]

It definitely sounds like an impulse question to me, but the tricky part seems to be finding the change in velocity. I got an answer of 1800N, but I'm not sure if I did it right (and don't want to confuse matters by writing out my approach if I'm wrong! ). Do you know what the right answer is?

 

[LBT]

In fact, the right answer is 2200 N. (4400 N for 2 feet, so 2200 N from one foot)

But I really don't know how they get this number.

I tried this:

F*t = delta p
F*t = mvf - mvi
but the mvf is a vector inverse to mvi, so
F*t = -mvf - mvi
F*t = -m(vf - vi)
to find mvf (at the ground), I used vf*vf = 2a*deltah ; vf = 4.47 m/s
F*t = -m(vf - vi)
F = -m(vf - vi)/t = -80kg*(4.47m/s - 0m/s)/0.2s = 1788 N (near 1800 N)
so 900 N from one foot.
BUT IT IS NOT THE RIGHT ANSWER...

did you have the same reasoning?

 

[prophecygirl]

Yep, same reasoning. It didn't seem quite right to me, but I couldn't think of another way to figure out the change in velocity. Sorry not to be more help -- maybe someone else will have a better approach! BTW, where is that question from? EK?

 

[Fort]

I got 2200 N on each foot.

Try using F = delta momentum/delta time.

It's an equation that gives the force on an object should the object change its momentum. So yeah, it's an impulse question. This question is pretty tricky, because the equation I just gave you calculates the net force, not necessarily the force exerted by his foot. So you need to take into account the effects of gravity.

Edit: where are you getting these questions; are you using a test prep?

 

[LBT]

Ok...

That is not the equation I just used?
F = delta momentum / delta time is pretty the same thing than F*t = delta p ...

How did you insert the gravity in this equation?

thank you

That's a question from Mcat physics 2009-2010 edition (kaplan)

 

[Fort]

Remember that the equation solves for the net force, not the force exerted by the ground on his feet; try drawing a free body diagram.

 

[prophecygirl]

EDIT: Nevermind -- I think I figured it out.

 

[LBT]

Nop!

I just dont get it!

F = delta momentum / delta time
(impulsion - ma) = (mvf - mvi)/t
impulsion = (80*4,47 - 0)/0.2 + 80*10
impulsion = 2588 for both feet

impulsion = 1294 for one foot.

What is wrong with my answer???

 

[Fort]

Nop!

I just dont get it!

F = delta momentum / delta time
(impulsion - ma) = (mvf - mvi)/t
impulsion = (80*4,47 - 0)/0.2 + 80*10
impulsion = 2588 for both feet

impulsion = 1294 for one foot.

What is wrong with my answer???

The net force on the jumper is

F(net) = delta momentum/delta time

F(net) = mv_f - mv_i /time

To calculate the velocity the moment before he hits the ground, we can use conservation of energy

mgh=1/2mv^2

v = sqrt(2gh)
v = sqrt(2*10*1)
v = sqrt(20) ~4.5

Now this is the velocity the moment he makes contact with the ground, it's also the velocity immediately after he jumps into the air because he returns to a height of 1 m.

Now we can use our force equation to calculate the net force exerted on the jumper. We'll define up as positive, so the velocity just before landing is negative.

F(net) = [80kg*(4.5 m/s -(-4.5m/s)]/0.2s

F(net) = 3600 N

Now that we have the net force, we can calculate the force exerted by a single foot.

F(net) = Force(feet) - Force(gravity)

=>

Force(feet) = F(net) + Force(gravity)

Force(feet) = 3600N + mg = 3600N + (80 kg)(10m/s^2)

Force (feet) = 3600N + 800N = 4400N upwards. This is the force exerted by his feet.

For a single foot, you just divide the value exerted by both feet in half. So you end up with 2200N.

Hope this helps.