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Like Tony Stark

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- Homework Statement
- Two bodies of equal mass are connected by an ideal rope and are inside a room which can accelerate vertically. The second body is attached to a spring of constant ##200 \frac{N}{m}##. Find the deformation of the spring when

A) the room moves with constant velocity

B) the room is accelerating upwards with ##2 \frac{m}{s^2}##.

C) the room is accelerating downwards with ##2 \frac{m}{s^2}##.

- Relevant Equations
- Newton's equations

I've solved all the cases in the non inertial system.

A) For ##m_1## we have

##x) P_{1x} -T=m.a_x##

##y) N_1 -P_{1y}=m.a_y##

For ##m_2## we have

##y) T+F_e -P_2=m.a_y##

As it moves with constant velocity I solve it setting ##a_x=0##. So for ##m_1## ##mgsin(\alpha)=T##, then I replace it in the tension of ##m_2## and find ##\Delta x##

B) For ##m_1##

##x) mgsin(\alpha)-f*.sin(\alpha)-T=0##

For ##m_2##

##y) Fe + T -mg-f*=0##

Where ##f*=2m##

Then, I replaced the tension from ##m_1## in ##m_2## and found the deformation. (As you see, I considered the elastic force pointing upwards in this case)

C) the same for ##B## just changing the sign of ##f*## and the elastic force.Is this right? Because I had some doubts about the sign of the elastic force.

A) For ##m_1## we have

##x) P_{1x} -T=m.a_x##

##y) N_1 -P_{1y}=m.a_y##

For ##m_2## we have

##y) T+F_e -P_2=m.a_y##

As it moves with constant velocity I solve it setting ##a_x=0##. So for ##m_1## ##mgsin(\alpha)=T##, then I replace it in the tension of ##m_2## and find ##\Delta x##

B) For ##m_1##

##x) mgsin(\alpha)-f*.sin(\alpha)-T=0##

For ##m_2##

##y) Fe + T -mg-f*=0##

Where ##f*=2m##

Then, I replaced the tension from ##m_1## in ##m_2## and found the deformation. (As you see, I considered the elastic force pointing upwards in this case)

C) the same for ##B## just changing the sign of ##f*## and the elastic force.Is this right? Because I had some doubts about the sign of the elastic force.

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