## physics problem

atalesfkxu
Posts: 1
Joined: Sun Sep 02, 2018 12:46 am

### physics problem

Hey, here is the problem I am working on

Some baseball pitchers are capable of throwing a fast ball at 100 mi/hr (45 m/s). The pitcher
achieves this speed by moving his arm through a distance of 1.5 m. Determine the average
net force that must be exerted on the 0.15 kg ball during the pitch.

I could do this using kinmatics but I need to do it with energy and work. The work on the ball is W=F*d*cos(t) but I'm not given an angle. Is it just zero degrees?

Also how is energy conserved if the ball gets KE? Do we assume it falls down so loses PE?

jeff
Posts: 39
Joined: Mon Mar 05, 2018 11:54 pm

### Re: physics problem

Hi atalesfkxu ,
atalesfkxu wrote:
Sun Sep 02, 2018 12:46 am
Hey, here is the problem I am working on

Some baseball pitchers are capable of throwing a fast ball at 100 mi/hr (45 m/s). The pitcher
achieves this speed by moving his arm through a distance of 1.5 m. Determine the average
net force that must be exerted on the 0.15 kg ball during the pitch.

I could do this using kinmatics but I need to do it with energy and work. The work on the ball is W=F*d*cos(t) but I'm not given an angle. Is it just zero degrees?
Yes, that's right! We assume here that the force on the ball is the same as the direction of the ball's motion, so that the angle between the is zero degrees.

Also how is energy conserved if the ball gets KE? Do we assume it falls down so loses PE?
That's a good question! It can be confusing to understand how to apply energy conservation to a single object.

But if we focus on just a single object, the idea is that its energy does not have to be conserved. Its energy is able to increase or decrease. The rule is that we have to look at the work being done. If any work is being done by non-conservative forces, then energy will not be conserved (i.e., it will change).

The force applied by the pitcher is non-conservative, so the work it does on the ball will change the ball's energy.

I hope this answered your question! If not, let me know and we can discuss it more.