What did I do wrong? 2 tests questions ( work and force questions )

[polarbearkids]

Homework Statement

1 question: How much work must be done to launch a 100-kg object to a height of 2 x 10^3 kilometer above the Earth's surface?

2.A person with the mass of 10 kg jumps up and down on her mattress with a spring constant of 4.3 nK.m. She does this by applying a sinusoidal force at a frequency of 1 Hz. Assuming no dampening force exists, if her amplitude is 3 cm, what is the maximum force she places on the mattress?

by the way, these are the only one I completely missed. He took off the maximum amount : ( And I still have no idea how to solve them. So I'm just wondering.

The Attempt at a Solution

1 question ( my attempt ) - okay, first I thought I would use the Vesc equation, but I simply got velocity out of it and I didn't know how to go from velocity to work. Then I had the idea, I would simply use the fact work is independent of the path and work = the change in potential energy
so mgh = work thus, 100kg x 9.8m/s^2 x 2x10^6 mThus, I got 1.96 x 10^9 joules. Now, I knew this was too simply to be right, but I am still confused why this is the case. Why can't I use the fact work equals the change in potential energy in this case?

2 question: I used the equation A = (Fsmall o / m) /( w^2 - wsmall o ^2) and solved for w^2. then used the fact the max acceleration = Aw^2. Then to get the force. I simply used F=ma. and got 13N
So, why is this wrong? I'm pretty sure, I converted to all the correct units. But, again, I thought this was too easy. Did I use an equation that wasn't applicable?

Thanks in advance!

 

[Enigman]

Homework Statement

1 question: How much work must be done to launch a 100-kg object to a height of 2 x 10^3 kilometer above the Earth's surface?

2.A person with the mass of 10 kg jumps up and down on her mattress with a spring constant of 4.3 nK.m. She does this by applying a sinusoidal force at a frequency of 1 Hz. Assuming no dampening force exists, if her amplitude is 3 cm, what is the maximum force she places on the mattress?

by the way, these are the only one I completely missed. He took off the maximum amount : ( And I still have no idea how to solve them. So I'm just wondering.

The Attempt at a Solution

1 question ( my attempt ) - okay, first I thought I would use the Vesc equation, but I simply got velocity out of it and I didn't know how to go from velocity to work. Then I had the idea, I would simply use the fact work is independent of the path and work = the change in potential energy
so mgh = work thus, 100kg x 9.8m/s^2 x 2x10^6 mThus, I got 1.96 x 10^9 joules. Now, I knew this was too simply to be right, but I am still confused why this is the case. Why can't I use the fact work equals the change in potential energy in this case?

P.E.=mgΔh is only valid for the approximation that $$Δh<<R_{earth}$$
You need to use $ΔU=-Gm1m2/R^2$

 

[rude man]

2.A person with the mass of 10 kg jumps up and down on her mattress with a spring constant of 4.3 nK.m.

What kind of spring constant is that? Do you mean 4.3 N/m??

 

[Chestermiller]

If the amplitude is 3 cm, and the frequency is 1 Hz, then represent the displacement at time t by:

x = 3 sin(2πt) (cm)
With this representation of the displacement, what is the velocity v in cm/sec?
What is the acceleration a in cm/sec²?

 

[polarbearkids]

What kind of spring constant is that? Do you mean 4.3 N/m??

Yup. sorry! kN/m kiloNewtons per meter