What is the y-component of the force by the floor on her during the collision?

  • #1
ryancerealbox
8
0
Homework Statement
Your lab partner jumps off a lab table and lands on the floor. Using video analysis, you measure her y-velocity during a certain time interval during the collision with the floor and graph 𝑝𝑦(𝑡) as shown below. Her mass is 63 kg. The slope of the best-fit line is 2625 kg·m/s2.

https://education.wiley.com/prod/w660d7fcb-fe26-43e3-9f17-93c3fbe62310/art/qb/qu/ch0/EAT_1593581387456_1_8615938273917383/qu/ch0/EAT_15932383140560_8051448095064999.png

2. What is the y-component of the force by the floor on her during the collision?

3. How does the force by the floor on her during the collision compare to the force by Earth?
Relevant Equations
|F floor| / |F gravity|
A previous question I got correct was: During this time interval, what is the y-component of the average net force on her? It was just 2625, and I thought question 2 would be the same answer but it wasn't. Not sure what to do.
 
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  • #2
I'm a bit confused about what's being measured by what here? ##kg\cdot m/s^2## is newtons, but how do you measure force by video analysis? Also, the floor is the Earth for all practical purposes.
 
  • #3
Sorry, forgot to include the actual graph.
1706298365676.png
 
  • #4
ryancerealbox said:
[...]
2. What is the y-component of the force by the floor on her during the collision?
[...]
During this time interval, what is the y-component of the average net force on her? It was just 2625, and I thought question 2 would be the same answer but it wasn't. Not sure what to do.
One question asks for the net. The other question asks for one of the contributions to the net. Can you list the forces that act on her during the collision?
 
  • #5
jbriggs444 said:
One question asks for the net. The other question asks for one of the contributions to the net. Can you list the forces that act on her during the collision?
The floor's opposite reaction and gravity? So 2625 and -9.8?
 
  • #6
ryancerealbox said:
The floor's opposite reaction and gravity? So 2625 and -9.8?
Be careful. One of those numbers is a force. The other is an acceleration.

Being careful in this case might include putting units after your numbers. That could help keep you on the straight and narrow.

Note that 2625 [Newtons] is the net force, not the normal force.
 
  • #7
jbriggs444 said:
Be careful. One of those numbers is a force. The other is an acceleration. But yes, you have identified the two relevant forces.

Being careful in this case might include putting units after your numbers. That could help keep you on the straight and narrow.
Since gravity is measured in m/s^2 its acceleration then, so the only force would be the 2625 kg*m/s^2? When I put that in it was wrong though
 
  • #8
ryancerealbox said:
Since gravity is measured in m/s^2 its acceleration then, so the only force would be the 2625 kg*m/s^2? When I put that in it was wrong though
No, you had the right idea in post #5 but you need two forces, not a force and an acceleration.
 
  • #9
So I multiply -9.8 by the mass 63 to get 617.4 kg*m/s^2 then subtract that from 2625 to get 2007.6 as the total force
 
  • #10
ryancerealbox said:
So I multiply -9.8 by the mass 63 to get 617.4 kg*m/s^2 then subtract that from 2625 to get 2007.6 as the total force
I thought 2625 was the total force. Q2 asks for the force from the floor only.
 
  • #11
ryancerealbox said:
So I multiply -9.8 by the mass 63 to get 617.4 kg*m/s^2 then subtract that from 2625 to get 2007.6 as the total force
If you manipulate the numbers enough times, you must get the right answer eventually!
 
  • #12
haruspex said:
I thought 2625 was the total force. Q2 asks for the force from the floor only.
My bad, so 2007.6 would be just the force from the floor, but that isn't the answer either.
 
  • #13
ryancerealbox said:
My bad, so 2007.6 would be just the force from the floor, but that isn't the answer either.
Write the equation relating the two forces to the net force.
 
  • #14
haruspex said:
Write the equation relating the two forces to the net force.
2625 kg*m/s^2 = 2007.6 kg*m/s^2 + 617.4 kg*m/s^2
 
  • #15
Isn't force a vector?
 
  • #16
I forgot about the negative, thanks guys it was 3242
 

1. What is the y-component of the force by the floor on her during the collision?

The y-component of the force by the floor on her during the collision is the component of the force that acts in the vertical direction, perpendicular to the floor. This component of the force helps determine the impact of the collision on her body in the vertical direction.

2. How is the y-component of the force calculated during a collision?

The y-component of the force during a collision can be calculated using Newton's second law, which states that the force is equal to the mass of the object multiplied by its acceleration. By analyzing the motion and forces acting on the object during the collision, the y-component of the force can be determined.

3. Why is the y-component of the force important during a collision?

The y-component of the force during a collision is important because it helps determine the impact of the collision on the object in the vertical direction. Understanding the y-component of the force can provide valuable information about the forces acting on the object and the resulting motion.

4. How does the y-component of the force by the floor on her affect her body during the collision?

The y-component of the force by the floor on her during the collision affects her body by exerting a force in the vertical direction. This force can cause her body to accelerate or decelerate vertically, depending on the magnitude and direction of the force. Understanding this component of the force is crucial for analyzing the effects of the collision on her body.

5. Can the y-component of the force by the floor on her be negative during a collision?

Yes, the y-component of the force by the floor on her can be negative during a collision if the force is acting in the opposite direction to the positive y-axis. A negative y-component of the force indicates that the force is acting downwards, causing her body to decelerate in the vertical direction.

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