Calculate Mechanical Energy & Heights of a Falling Stone - Physics Homework Help

[Roro312]

Homework Statement

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

Homework Equations

Emech= Eg+ Ek

Ek= 1/2mv^2

Eg= mgh

The Attempt at a Solution

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

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


the rest i don't get ...im soo confused and i have a test tomorrow
please help me

 

[zachzach]

Homework Statement

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

Homework Equations

Emech= Eg+ Ek

Ek= 1/2mv^2

Eg= mgh

The Attempt at a Solution

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

b) Ek= (.2)(9.8)(1.5)
=1.47Jthe rest i don't get ...im soo confused and i have a test tomorrow
please help me

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

Conservation of energy [tex] E_i = E_f [/tex]

(a) [tex] U =mgh [/tex]

Looks right.

(b) [tex] KE = \frac{1}{2}mv^2 [/tex]

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.

 

[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

 

[zachzach]

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

(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.

 

[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?

 

[zachzach]

(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?

I got the same for both.

 

[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 wouldn't have got it.. THANKS ALOT:)


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

 

[zachzach]

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 wouldn't have got it.. THANKS ALOT:)


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

(e) Yes

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

 

[Roro312]

Thanks a lot for your help Zach... I appreciate it

 

[zachzach]

No prob.