# Mechanical energy

1. Apr 28, 2010

### Roro312

1. The problem statement, all variables and given/known data

A bridge is 25 m above a road below. If a .20kg stone is thrown upward from the bridge at 22.0m/s, calculate:

a) the gravitational energy of the stone as it is thrown upwar
b) the kinetic energy of the stone as it is thrown up
c) the maximum height (above the ground) achieved by the stone
d) the speed of the stone as it hits the road
e) the total mechanical energy of the stone as it falls pat the bridge on the way down

2. Relevant equations

Emech= Eg+ Ek

Ek= 1/2mv^2

Eg= mgh

3. The attempt at a solution

a) Eg= (.20)(9.8)(25)
= 49J

b) Ek= (.2)(9.8)(1.5)
=1.47J

the rest i dont get ...im soo confused and i have a test tomorrow

2. Apr 28, 2010

### zachzach

So set the zero point of potential energy on the ground. Then:

Conservation of energy $$E_i = E_f$$

(a) $$U =mgh$$

Looks right.

(b) $$KE = \frac{1}{2}mv^2$$

Where did you get 1.5?

(c) At the maximum height v = 0, so all of the initial kinetic energy was converted into potential energy.

(d) At the ground h = 0 so all of the energy has been converted into kinetic energy.

(e) Mechanical energy is conserved.

3. Apr 28, 2010

### Roro312

ok so for b) i got 2371.6J

but for c) how do we change initial energy to potentional energy..

Thanks again for ur response... i greatly appreciate it:D

4. Apr 28, 2010

### zachzach

(b) Try the calculation again.

(c) So initially it is going 22 m/s, this gives it kinetic energy (the amount which is the answer to b). It reaches its maximum height when all of this kinetic energy is converted into potential energy (set equal). This will only give you the height above the bridge though so do not forget to add the height of the bridge as well.

5. Apr 28, 2010

### Roro312

(b) 48.4J

(c) 1/2 v2 = gh
½ (22.0)2 = 9.8h
½ (484) = 9.8h
242 = 9.8h
9.8 9.8

24.69 = h

Therefore the max height equals 25 m + the height of the bridge equals 50 m

is that right?

6. Apr 28, 2010

### zachzach

I got the same for both.

7. Apr 28, 2010

### Roro312

d) i got 22m/s

e)Mmech=(.2)(9.8)(25)+1/2(.2)(22)^2
=97.4 J

o my gosh.. i know i bugged you soo much but seriously without your help i wouldnt have got it.. THANKS ALOT:)

just the last two... did you get the same?

8. Apr 28, 2010

### zachzach

(e) Yes

(d) No what I got. All the energy is now kinetic.

9. Apr 28, 2010

### Roro312

Thanks alot for your help Zach... I appreciate it

10. Apr 28, 2010

No prob.