Kinetic Energy,Work Energy therom, Maximum Height

[Dejahboi]

Homework Statement

A 0.145-kg baseball is struck by a bat 1.20 m above the ground, popping straight up at 21.8 m/s. (a) What's the ball's kinetic energy when it leaves the bat? (b) How much work is done by gravity once the ball reaches maximum height? (c) Use your answer in part (b) to find that maximum height. (d) Find the work gravity does on the ball from when it's batted until it hits the ground. (e) Ignoring air resistance, use your answer in part (d) to find the ball's speed at the ground.

Homework Equations

K= 1/2mv²

The Attempt at a Solution

a) I used K=1/2mv²

so,
(.5)(.145 kg)(21.8 m/s)²= 34.45 J

b) I'm not sure how I did it but I got = -34.45 J

c)According to the free body diagram I drew, my initial velocity is 21.8 m/s, and V_f is 0. I'm not sure what equation to use here, I used W_g= -mgΔy but my answer does not match the book's answer.

As for the rest of the problem, I have no clue how to do it.

 

[PhanthomJay]

Homework Statement

A 0.145-kg baseball is struck by a bat 1.20 m above the ground, popping straight up at 21.8 m/s. (a) What's the ball's kinetic energy when it leaves the bat? (b) How much work is done by gravity once the ball reaches maximum height? (c) Use your answer in part (b) to find that maximum height. (d) Find the work gravity does on the ball from when it's batted until it hits the ground. (e) Ignoring air resistance, use your answer in part (d) to find the ball's speed at the ground.

Homework Equations

K= 1/2mv²

The Attempt at a Solution

a) I used K=1/2mv²

so,
(.5)(.145 kg)(21.8 m/s)²= 34.45 J

yes

b) I'm not sure how I did it but I got = -34.45 J

well you did something right or copied the book answer

c)According to the free body diagram I drew, my initial velocity is 21.8 m/s, and V_f is 0. I'm not sure what equation to use here, I used W_g= -mgΔy but my answer does not match the book's answer.

I'm not sure why it doesn't match...show your work

As for the rest of the problem, I have no clue how to do it.

Well for (d) you got (b) right somehow on the upward journey, so you should be able to find the work done by gravity on the downward trip, then add them up . For (e), use the work-energy theorem, and show your work.

 

[Dejahboi]

For (c)

I set -34.45 J = (-0.145 kg) (9.8 N/m) H

So,

-34.45/ -1.421 = 24.24 m

Book says 25.45 m but I can be doing something really wrong.

 

[Simon Bridge]

(b)When the ball has reached the top of it's trajectory, what is it's kinetic energy? How is work related to kinetic energy?

(c) how does your answer and the book's differ?

(d) is like (b) ... what is the KE it started with? Ended with? What is the relationship between KE and work?

(e) follows from (d) ... you have the final KE and the relationship between v and KE.

 

[Dejahboi]

(b)When the ball has reached the top of it's trajectory, what is it's kinetic energy? How is work related to kinetic energy?

W= (1/2)kx²?
K = 1/2mx²

(c) how does your answer and the book's differ?

Well I got 24.24 m and the book has 25.45 m

 

[Simon Bridge]

W= (1/2)kx2?

That would be energy stored in a spring ... what is the realtionship between work and kinetic energy? That would be the work-energy theorem PhantomJay was talking about.

K = 1/2mx2

? Clearly you don't know - you can look it up: but here's a quickee refresher:

Kinetic energy: $K=\frac{1}{2}mv^2$

Work-Energy: $W=\Delta E$ ... work is change in energy.

The work done on something is the change in it's kinetic energy.

So, in terms of kinetic energy: $W=K_f-K_i$

Well I got 24.24 m and the book has 25.45 m

So the book had a height 1.2m higher than yours ... hmmm... isn't that a familiar figure? Can you find that number in the question?

 

[Dejahboi]

That would be energy stored in a spring ... what is the realtionship between work and kinetic energy? That would be the work-energy theorem PhantomJay was talking about.? Clearly you don't know - you can look it up: but here's a quickee refresher:

Kinetic energy: $K=\frac{1}{2}mv^2$

Work-Energy: $W=\Delta E$ ... work is change in energy.

The work done on something is the change in it's kinetic energy.

So, in terms of kinetic energy: $W=K_f-K_i$

So the book had a height 1.2m higher than yours ... hmmm... isn't that a familiar figure? Can you find that number in the question?

Yeah, of course I wouldn't know, that's why I asked for help. Anyways, I have a list of equation I written down from class, but the professor had a tiny mistake categorizing the work of spring and the work energy, and yeah I have

ΔK = W

Now I lost track of what I'm trying to solve for :/..

 

[Simon Bridge]

Yeah, of course I wouldn't know, that's why I asked for help.

No worries - sometimes people ask a question not realizing they know the relationships to use.

Anyways, I have a list of equation I written down from class, but the professor had a tiny mistake categorizing the work of spring and the work energy

That won't be entirely a mistake - the idea is that the list of equations is to act to jog your memory ... you are expected to learn the physics and let that knowledge guide your choice. eg. you were expected to realize that the little "k" normally appears with an "x" if it is the spring constant. Just like it appears with a q or a Q if it is the coulomb constant and with a T if it is the Boltzman constant. You needed something with a W and a K or an E in it.

Don't worry - it is common to need to reread the question after thrashing out concepts ;)