What is the average force exerted on a baseball during a pitch?

In summary, a pitcher can accelerate a 0.15 kg ball from rest to 90 mi/h in a distance of 1.6 m. To find the magnitude of the average force exerted on the ball during the pitch, you need to use the equations a=v/t and xf=xi+vixt+1/2(a)t^2. The initial velocity of the ball is 0 since it starts at rest. Convert the final speed into m/s and use it to obtain an expression for the acceleration, which can then be used to calculate the net force using Fnet=ma. It may also be helpful to memorize the equations vf^2=vi^2+2ad for constant acceleration problems.
  • #1
XPX1
55
0

Homework Statement



In baseball, a pitcher can accelerate a 0.15 kg ball from rest to 90 mi/h in a distance of 1.6 m.
(a) What is the magnitude of the average force exerted on the ball during the pitch?
____N



Homework Equations


a=d/t
xf=xi+vixt+1/2(a)t^2
Fnet=ma


The Attempt at a Solution



I know that to find out this problem I need to find out the Fnet. I know that Fnet=ma but how do I find out the a!?

I know to figure out acceleration its distance/time but..

How do I figure out acceleration? This is getting really frustrating. It gets to 90 mi/hr in 1.6m.

Does mass have anything to do with finding out the acceleration? Please help!
 
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  • #2
XPX1 said:
I know to figure out acceleration its distance/time but..

How do I figure out acceleration? This is getting really frustrating. It gets to 90 mi/hr in 1.6m.

Does mass have anything to do with finding out the acceleration? Please help!

Acceleration is *not* distance/time. Velocity is distance/time, whereas acceleration is rate of change of velocity (i.e. velocity/time)
 
  • #3
Also, what does it mean by (a) what is the magnitude of the average force exerted on the ball during the pitch!?
 
  • #4
I don't know what vi=?

is vi=90? or is that vf?
 
  • #5
XPX1 said:
Also, what does it mean by (a) what is the magnitude of the average force exerted on the ball during the pitch!?
The magnitude is simply the size of the force exerted on the ball; [itex]|\vec{F}| = m|\vec{a}|[/itex]
XPX1 said:
I don't know what vi=?
is vi=90? or is that vf?
Vi is the intial velocity, so yes, vi=90mi/h. It would however, be sensible to convert this speed into m/s.
 
  • #6
Any help on how to plug this in? Now that I have the velocity, I need to find the time, dang these problems are confusing, how do I figure out the time!?
 
Last edited:
  • #7
any ideas?
 
  • #8
I have

vi=90 mi/hr
mass=0.15
xf=1.6

I can't do anything with this information! I've tried velocity versus time graphs, position versus time graphs, everything! It seems like this problem does not give me enough information to solve it!
 
  • #9
Change vi

I'm sorry but vi is not equal to 90 mi/hr.

The initial velocity is vi=0 (the key word is at rest).

vf=90 mi/hr but this is useless until you convert to m/s.

Now you know vi, vf, xf, xi=0, so plug into your equation, do some algebra to isolate acceleration and this is easily solved.


PS. later you'll learn work-energy and it'll save you a few steps, but for now the basics.
 
  • #10
But.. How can I solve for acceleration if I can't solve for time!? Please help!
 
  • #11
XPX1 said:
But.. How can I solve for acceleration if I can't solve for time!? Please help!

v = at, as I am sure you know. Well, it's v = v0 + at. But your v0 = 0.

You can use this to eliminate the 't' in the distance equation.

Dorothy
 
  • #12
That dosn't make any sense to me!

v0 = 0... Ok

v = 0+at


now how do I figure out acceleration and time!? Can somebody please do an equation of this so I can see how this works out, I've been trying to figure out this problem for 3 days now, and I've had no luck.
 
  • #13
XPX1 said:

Homework Equations


a=d/t
xf=xi+vixt+1/2(a)t^2
Fnet=ma

I don't know whether you've finished this question yet, but I can see that your last post hasn't been answered.

Well, to start with, note that your first equation is incorrect, it should read a=v/t, and your second eqn should read xf=xi+vi*t+1/2(at^2).

Now, before calculating anything, convert the final speed vf, into m/s.

Using your first equation, obtain an expression for the acceleration, noting that vi=0, so a=vf/t, which gives t=vf/a.

Now, use the second eqn, with xi=0, vi=0, and xf=1.6m. This gives 1.6=1/2(at^2), but from above, t=vf/a, and so we can substitute this into the expression to yield 1.6=(1/2)*a*vf^2/a^2.

From here, you can calculate a (on substituting in the converted vf) and thus go on to calculate the net force, using your third equation.
 
  • #14
Listen to cristo but add in one short cut. If you grab a physics book you will find an equation with no time. It looks something like this.

vf(squared)=vi(squared)+2ad
where v-velocity, a-acceleration and d-distance from there this should be easy.

On the last note mechanics is much easier if all 4-equations of motion are memorized for constant acceleration.
 
  • #15
I'd definitely agree, you should memorise these equations so you can have them at your fingertips when attempting this sort of question.
 

1. What is Fnet acceleration problem?

The Fnet acceleration problem is a physics concept that involves calculating the net force acting on an object and using that information to determine the object's acceleration. It is based on Newton's second law of motion, which states that the net force on an object is directly proportional to its acceleration and inversely proportional to its mass.

2. How do you calculate Fnet acceleration?

To calculate Fnet acceleration, you first need to identify all the forces acting on the object. Then, you can use the formula Fnet = ma, where Fnet is the net force, m is the mass of the object, and a is the acceleration. If there are multiple forces acting on the object, you will need to add them together to find the net force.

3. What are some common examples of Fnet acceleration problems?

Some common examples of Fnet acceleration problems include objects sliding down a ramp, a ball being thrown into the air, and a car accelerating on a highway. In each of these situations, there are multiple forces acting on the object, and the net force determines the object's acceleration.

4. How does the direction of forces affect Fnet acceleration?

The direction of forces can greatly affect Fnet acceleration. If the forces are acting in the same direction, the net force will be larger and the acceleration will be greater. If the forces are acting in opposite directions, the net force will be smaller and the acceleration will be less. Additionally, the direction of the net force will determine the direction of the object's acceleration.

5. How is Fnet acceleration related to motion?

Fnet acceleration is directly related to an object's motion. According to Newton's second law, the greater the net force, the greater the acceleration, and therefore the greater the change in motion. This means that a larger net force will cause an object to accelerate more quickly and reach a higher velocity, while a smaller net force will result in slower acceleration and lower velocity.

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