# Kinetic E & Potential E Question

1. Jun 2, 2010

### physics(L)10

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

A 55kg mass is projected vertically with and initial speed of 30m/sec
a) what is original kinetic energy
b) What is kinetic E after 4.5 seconds
c) What is chg in gravitational potential E in these 4.5seconds?

2. Relevant equations

W=mv2/2
W=(m/2)(gt)2
W=mgh

3. The attempt at a solution

a) Original =(55)(30)2/2

b) W=(m/2)(gt)2

=(55/2)[(9.8)(4.5)]2

c) W=mgh

=(55)(9.8)h

2. Jun 2, 2010

### Staff: Mentor

This is correct.

This is not correct. Rethink how to calculate the speed; it's not equal to gt. (It's slowing down, for one thing. The acceleration is negative.)

3. Jun 2, 2010

### physics(L)10

ok I think I know,

v2=v1 -at
0=55-(9.8)(4.5)

Is this correct?

And how about c? you didn't comment on that

4. Jun 2, 2010

### Staff: Mentor

Good. (I assume you mean v2 = ..., not 0 = ...)

You need the answer to b to get c. (You aren't given the height.)

5. Jun 2, 2010

### physics(L)10

Yes my bad. and for c, does it look something like this?:

mgh=mv2/2

The m's cancel and you get:

gh=v2/2

Solving for h,

h=v2/2g

And then you could work out W=mgh?

6. Jun 2, 2010

### Staff: Mentor

Not exactly. Use conservation of energy like this:
KE1 + PE1 = KE2 + PE2
Then solve for the change in PE.

(Your equation assumes that the final KE is zero, which is not the case here.)

Note that there's no need to solve for h, just use energy conservation.

7. Jun 2, 2010

### physics(L)10

So let's say, initial speed = 55 and final = 25. These would be the only differences in KE1,KE2?

8. Jun 2, 2010

### Staff: Mentor

Those would be the speeds, from which you'd calculate KE1 and KE2 and then the difference.

9. Jun 2, 2010

### physics(L)10

How would you find out the height difference to find the potential energy difference?

10. Jun 2, 2010

### Staff: Mentor

You do not need the height difference. They ask for the change in PE, which can be found using conservation of energy.

If you wanted to find the height difference, use kinematics. But that's not necessary.

11. Jun 2, 2010

### physics(L)10

Would this work? I find change in KE which then can be used in the equation -KE=PE?

12. Jun 2, 2010

### Staff: Mentor

I think you mean -ΔKE = ΔPE. Sure that will work--that's equivalent to the equation I gave in post #6.

ΔKE = KE2 - KE1
ΔPE = PE2 - PE1

13. Jun 2, 2010

### physics(L)10

Okay I think I got it. Thank you :D :D