What Is the Average Force Exerted by the Floor on a Jumping Lizard?

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Homework Help Overview

The discussion centers around calculating the average force exerted by the floor on a lizard as it jumps. The problem involves concepts from mechanics, specifically relating to forces, energy, and motion. Participants explore the dynamics of the lizard's jump, including its mass, height changes, and the forces acting during the jump.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss using kinematics to determine the final speed of the lizard's center of mass at the point of losing contact with the floor. There are attempts to relate kinetic energy to the work done by the floor's force. Some participants express confusion about the relationship between distance, time, and velocity, while others suggest breaking down the problem into smaller steps.

Discussion Status

There is an ongoing exploration of different methods to calculate the forces involved. Some participants have provided calculations and reasoning, while others have questioned the accuracy of their results. The discussion reflects a collaborative effort to clarify concepts and verify calculations without reaching a definitive conclusion.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is also a focus on understanding the forces during the brief contact time between the lizard and the floor.

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[SOLVED] force in a non-isolated system

I don't normally do something like this, but I'm stuck on my homework.

A lizard jumps, pushing with its legs against the ground. Assume the lizard's inertia is .05 Kg. Its center of mass starts out .5 cm above the floor, and raises 1 cm to 1.5 cm while it is pushing against the floor. At that height, its feet lose contact with the floor and it continues to travel up to a height of .15 m. What is the magnitude of the average force exerted by the floor on the lizard?

I know that F=ma, but I have no idea how to find the acceleration without the time the action took. I also know that if I have to use the change in center of mass, I'll have to convert the cm to m. But other than that, I'm totally lost.

Any help would be appreciated.
 
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Start by finding the final speed of the center of mass just as the lizard loses contact with the floor. Hint: Use kinematics to find that speed, given the height it reaches.

Then you can figure out what the force must have been to produce such a change in kinetic energy. (Don't neglect the lizard's weight.)
 
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I know that I'm going to sound really stupid, but I still don't get it. I skipped this problem and moved on to others, and did what I could with those, but now I'm just more confused than ever when I look back at this one. :S
 
Just try it step by step. If you threw a ball straight up and it traveled a maximum distance d above your hand (before falling back down again), could you figure out how fast it must have been moving when it left your hand? Same idea here.

What's the change in height of the center of mass, from the point where the lizard loses contact with the floor to the point where it reaches maximum height?
 
I'm still lost. Every formula I know for finding velocity over some distance depends on the time it took to travel that distance.

The only solution I can come up with looks like this:

Vfinal=(2gh)^(1/2), where g=9.8 and h=.15-.015=.135

So, Vfinal=1.63 m/s

KE=1/2mv^2, where v=1.63 and m=.05 kg

So, KE=.07 J

This represents total work done and W=Force(change in distance), so .07=force(.135)

Force=.52 N

Is this right at all?
 
Kris1214 said:
I'm still lost. Every formula I know for finding velocity over some distance depends on the time it took to travel that distance.
You don't sound as lost as you think!

The only solution I can come up with looks like this:

Vfinal=(2gh)^(1/2), where g=9.8 and h=.15-.015=.135

So, Vfinal=1.63 m/s

KE=1/2mv^2, where v=1.63 and m=.05 kg

So, KE=.07 J
Sounds good to me! Although I would not round off at this point.

If we were a bit more clever, we could have found the KE directly and skipped finding the speed:
KE = mgh = 0.06615 J

This represents total work done and W=Force(change in distance), so .07=force(.135)

Force=.52 N

Is this right at all?
Since we're interested in the force of the floor on the lizard, we only care about what happens during the time the lizard is pushing on the floor. What distance does the center of mass move during that time? Use that to find the net force on the lizard, then analyze the forces to figure out what the force of the floor must have been.

You're doing good.
 
Ok, so...

KE during contact with floor=.05*9.8*.01=.06615 J

.06615=Force(.01); Force=.49 N during contact with the floor

This gives us a net force of 1.01 N on the lizard.

Since force due to gravity while the lizard is in contact with the floor=mg=.05(9.8)=.49 N.

And assuming that the downward force of gravity is negative, we get a total upward force of the floor on the lizard of 1.01-(-.49)=1.5 N.

Right, or did I just totally mess it up?
 
Kris1214 said:
Ok, so...

KE during contact with floor=.05*9.8*.01=.06615 J

.06615=Force(.01); Force=.49 N during contact with the floor

This gives us a net force of 1.01 N on the lizard.
Not sure where you got 1.01 N. You have the equation correct:
.06615=Force(.01)

Redo your solution for the net force. (.49 N is the weight of the lizard; you'll use that later.)
 
Sorry, must have hit the wrong button on the calculator.

Force while in contact with floor is 6.615 N, and therefore Net force is 7.135 N.
 
  • #10
The net force on the lizard is 6.615 N (upward), and therefore the force of the floor must be 6.615 N + mg.
 
  • #11
So, that gives us a force on the floor of 7.105 N.

The next part of the question asks about the acceleration of the lizard during the launch time so 7.105/mass=142.1 m/s^2. But doesn't this sound a little too fast?
 
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  • #12
Kris1214 said:
So, that gives us a force on the floor of 7.105 N.
OK.

The next part of the question asks about the acceleration of the lizard during the launch time so 7.105/mass=142.1 m/s^2.
The acceleration = (net force)/mass.
But doesn't this sound a little too fast?
The poor lizard needs to toss his body pretty high in the air (compared to his size), so he needs to produce a considerable acceleration to get himself up to speed in that short a distance.
 
  • #13
Ok, I think that I finally get it. No garuntee I'll be able to do a similar problem in the future. But thanks for all your help. I really appreciate it. :D
 

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