1. Sep 24, 2014

### liamke70

Just wondering if apparent weight affects kinetic energy or is it only mass that affects it regardless of apparent weight?

for example, a helicopter is falling but is providing a small thrust. say the helicopters mass is 50kg and it provides 100N of thrust upwards

therefore, using w=mg, the weight would be 490.5 N , and subtracting the 100 N of thrust would make the apparent weight 390.5

say its falling 50 meters, the Ep=mgh, Ep=390.5x50, Ep= 19525N

then, when using Ek=0.5mv^2, to find the velocity before it hits the ground, dose the apparent weight need to be considered or is it only the mass

would it be correct to say v=(2(50)(19525))^0.5=1397.32m/s or does the apparent weight need to be thought about when calculating velocity like this?

Thanks

2. Sep 24, 2014

### Staff: Mentor

To calculate the velocity before it hits the ground, you need to consider all the forces acting on it, including the thrust. The kinetic energy is given by $0.5mv^2$, where m is the mass.

3. Sep 24, 2014

### liamke70

how would i include upwards thrust?

4. Sep 24, 2014

### Staff: Mentor

You already did. The net force on the helicopter is gravity (mg) downward and thrust (100 N) upward. For some reason, you called that net force "apparent weight". I don't think that is standard terminology.

The work done on the helicopter as it falls will equal the net force X the height fallen. Just what you calculated. Which equals the increase in KE.

I would not use the term "apparent weight" in this context. What you calculated is not gravitational PE, mgh, but the net work done.

5. Sep 24, 2014

### liamke70

okay, so your saying the 100N of thrust doesnt alter the mass but just the net force, and then net force x h (energy) that becomes kinetic energy. so then can i then just put the energy(net force x h) and mass(50kg) into E= 1/2mv^2 to get the velocity?

Thank you very much for your replies

6. Sep 24, 2014

### Khashishi

Kinetic energy just depends on the mass and velocity, and not the force. The thrust doesn't change the "m" in the kinetic energy. Think about it: if upward thrust actually reduced the mass, then the kinetic energy would go to zero if the helicopter flew at a level elevation and would be negative if rising. That makes no sense.

7. Sep 24, 2014

### Staff: Mentor

Exactly.

8. Sep 24, 2014

### liamke70

Thank you :)

while i have you here, could you please answer one more thing?

if a spring is at its point of maximum deflection, does it the body the spring is attached to have any acceleration?

the body is landing on the springs

9. Sep 24, 2014

### Staff: Mentor

You tell me: What determines whether a body accelerates?

10. Sep 24, 2014

### liamke70

well theres no change in velocity over time so i guess theres none

i think i'm just paranoid that my lecturer is trying to trick me haha

11. Sep 24, 2014

### Staff: Mentor

You didn't answer my question! What determines whether a body accelerates? (Think Newton's laws.)

12. Sep 24, 2014

### liamke70

f=ma , a=f/m

forces and balances f=0

therefore a = 0

is that right?

13. Sep 24, 2014

### Staff: Mentor

If the net force is zero, then the acceleration is zero. So, if you gently rest a body on a spring, so it just stays there, you know the net force is zero and thus the acceleration is zero.

But if you drop the body on the spring, the spring will compress beyond the equilibrium point. Thus the body accelerates at that point of max compression. If you think about it, that should be obvious, since if it didn't accelerate it would just stay at max compression. But you know it will bounce back.

Don't confuse the fact that the body momentarily comes to rest with it having no acceleration. You can have zero velocity (for an instant) and still be accelerating.

14. Sep 24, 2014

### liamke70

okay so if energy causing the spring to compress say 6 J(Ep) and its maximum deflection is 0.02m then how would the acceleration be calculated?

maybe
x being deflection length
6= max
6/mx=a

15. Sep 24, 2014

### liamke70

or is it rather

kx/m=a

x= deforamtion
k= spring rate

16. Sep 24, 2014

### Staff: Mentor

Almost. You need the net force, not just the force from the spring. If you drop something onto the spring and it has a max compression of 'x', then the net force would be kx - mg. You'd use that net force to calculate the acceleration.

Often your task will be to first solve for the maximum compression. To do that you would use energy methods.

17. Sep 24, 2014

### liamke70

i have used energy methods(potential energy to work done to the spring) and gotta the max compression

so now what do i do to get the acceleration at maximum compression?

rearrange the net force (kx -mg) to get acceleration?

18. Sep 24, 2014

### Staff: Mentor

Good.

kx - mg is the net force; Newton's 2nd law will allow you to calculate the acceleration.

19. Sep 24, 2014

### liamke70

i dont see how i can use newtons second law f=ma to solve for acceleration in this case

a= f/m

cant be applied as im dealing with energy and work

20. Sep 24, 2014

### Staff: Mentor

You'd use energy and work in one part of the problem, then forces and Newton's 2nd law in another part. No problem. You know the net force, so nothing stops you from calculating the acceleration.