Q)The two blocks in an atwood machine have masses 2kg and 3 kg . find the work done by gravity during the fourth second after the system is released from rest?Q)A block of mass 100 g is moved with a speed of 5 m/s at the highest point in a closed circular tube of radius 10 cm in a vertical plane. The cross- section of the tube is such that the block just fits in it. The block makes several oscillations inside the tube and finally stops at the lowest point. Find the work done by the tube on the block during the process.Please give the answer by explaining the question and what should be our mindset at this questions. Thank you

[Q.1]

          By solving these simple force equations,

                                                                      T-2mg=2ma       

                                                                       3mg-T=3ma

                                                                        a=g/5

          this'll give distance moved by the blocks in 4th sec.=1/2*(g/5)*(4²-3²)=7mtrs.  [this is a case of constant acceleration ;S_nth=1/2a((t_n)²-(t_n-1)²]

Now 3kg. block moves down & 2kg. block moves up by the same 7 mtrs.So for 3kg. block  gravity do (+ve) work as force & displacement has same direction while for 2kg. block it is (-ve).Equivalently (3-2=1)kg. move down by 7mtrs. & so total work done by gravity=1*10*7=70 J(F.dr)

Here this much work done by the gravity is transferred to the system as increase in their total kinetic energy which you could find by applying(v=u+at & K.E.=1/2 mv²).What u've to keep in mind that in case of external conservative field present(gravity for the system here) you couldn't use total change in potential & kinetic energy(mechanical energy)=work done by external agency.To be safe you could always do the whole thing in most basic way as i've done the question here by using basic concepts & mathematical formula.But you should know the physics also!!!

Note:For more clarity & greater grasp on topic & physics in general consult "Resnick ,Hallidey & Walker "(i think chapter 7"Work & Energy" in 4th ed. ;i'm not sure though).

[Q.2]  Total change in mechanical enery of block=Total work done by frictional force(exerted by tube on the block)=(0.1*10*0.2)+(1/2*0.1*5²)=1.45J.

Here frictional force being non-conservative force the above concept is used.Always keep in mind for any system

1. Only external force will do any work.That force being conservative change in total mechanical energy will be zero.(work done=change in pot. energy=change in kin. energy)

2.For non-conservative force work done=Total change in mechanical energy.

As for mindset for any problem(in all 3 subjects & especially in physics) never fudge the basics.Concepts must be understood fully & unambiguously & practiced regularly.

Good Luck!