Kinetic E & Potential E Question

In summary, the 55kg mass is projected vertically with an initial speed of 30m/sec. After 4.5 seconds, its original kinetic energy is decreased by 9.8J due to gravitational potential energy.
  • #1
physics(L)10
101
0

Homework Statement



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?

Homework Equations



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


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
 
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  • #2
physics(L)10 said:
a) Original =(55)(30)2/2
This is correct.

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

=(55/2)[(9.8)(4.5)]2
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
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
physics(L)10 said:
ok I think I know,

v2=v1 -at
0=55-(9.8)(4.5)
Good. (I assume you mean v2 = ..., not 0 = ...)

And how about c? you didn't comment on that
You need the answer to b to get c. (You aren't given the height.)
 
  • #5
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
physics(L)10 said:
Yes my bad. and for c, does it look something like this?:

mgh=mv2/2
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
So let's say, initial speed = 55 and final = 25. These would be the only differences in KE1,KE2?
 
  • #8
physics(L)10 said:
So let's say, initial speed = 55 and final = 25. These would be the only differences in KE1,KE2?
Those would be the speeds, from which you'd calculate KE1 and KE2 and then the difference.
 
  • #9
How would you find out the height difference to find the potential energy difference?
 
  • #10
physics(L)10 said:
How would you find out the height difference to find the potential energy difference?
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
Would this work? I find change in KE which then can be used in the equation -KE=PE?
 
  • #12
physics(L)10 said:
Would this work? I find change in KE which then can be used in the equation -KE=PE?
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
Okay I think I got it. Thank you :D :D
 

1. What is the difference between kinetic energy and potential energy?

Kinetic energy is the energy possessed by a moving object, while potential energy is the energy stored in an object due to its position or state.

2. How is kinetic energy calculated?

Kinetic energy is calculated using the formula KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity.

3. What factors affect the amount of potential energy an object has?

The amount of potential energy an object has is affected by its mass, height, and the strength of the force acting on it.

4. Can energy be converted from kinetic to potential or vice versa?

Yes, energy can be converted from kinetic to potential or vice versa. For example, when an object is thrown upwards, its kinetic energy decreases as it reaches its highest point, and its potential energy increases.

5. How is the conservation of energy related to kinetic and potential energy?

The conservation of energy states that energy cannot be created or destroyed, only transferred or transformed. In the case of kinetic and potential energy, they can be converted into each other, but the total amount of energy in a system remains constant.

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