pulleys-mechanics, pls help i cant solve the whole question
Two particles A and B of mass 1kg and mkg respectively, where m>1 are attached to the ends of a light inextensible string which passes over a small fixed smooth pulley.The particles are released from rest and move with th string taut and vertical.
i) show that the acceleration of the system is equal to (m-1)g/m+1
ii) find the tention in the spring in terms of m and g expressing your answer as a single algebraic fraction in its simple form.
When the system is realeased from rest both particles are 52.5 cm above ground level and 60com below the level of the pulley.B hits the ground after half a second,
iii)fiind the value of m
iv) find the speed with which B hits the ground
i) show that the acceleration of the system is equal to (m-1)g/m+1
ii) find the tention in the spring in terms of m and g expressing your answer as a single algebraic fraction in its simple form.
When the system is realeased from rest both particles are 52.5 cm above ground level and 60com below the level of the pulley.B hits the ground after half a second,
iii)fiind the value of m
iv) find the speed with which B hits the ground
Original post by ugauafidufd
Two particles A and B of mass 1kg and mkg respectively, where m>1 are attached to the ends of a light inextensible string which passes over a small fixed smooth pulley.The particles are released from rest and move with th string taut and vertical.
i) show that the acceleration of the system is equal to (m-1)g/m+1
ii) find the tention in the spring in terms of m and g expressing your answer as a single algebraic fraction in its simple form.
When the system is realeased from rest both particles are 52.5 cm above ground level and 60com below the level of the pulley.B hits the ground after half a second,
iii)fiind the value of m
iv) find the speed with which B hits the ground
i) show that the acceleration of the system is equal to (m-1)g/m+1
ii) find the tention in the spring in terms of m and g expressing your answer as a single algebraic fraction in its simple form.
When the system is realeased from rest both particles are 52.5 cm above ground level and 60com below the level of the pulley.B hits the ground after half a second,
iii)fiind the value of m
iv) find the speed with which B hits the ground
Part (i) same method as the last question.
Part (ii) you can substitute your acceleration into either of the equations of motion for A or B to find the string tension.
Parts (iii) and (iv) you can use suvat.
Will the answer for ii be 2mg?
Original post by old_engineer
Part (i) same method as the last question.
Part (ii) you can substitute your acceleration into either of the equations of motion for A or B to find the string tension.
Parts (iii) and (iv) you can use suvat.
Part (ii) you can substitute your acceleration into either of the equations of motion for A or B to find the string tension.
Parts (iii) and (iv) you can use suvat.
how will I use suvat for iii
Original post by ugauafidufd
Will the answer for ii be 2mg?
No, that's not what I got. Probably best if you post your working.
Original post by ugauafidufd
how will I use suvat for iii
You have the values of "s" and "t", you know that u=0 and you have "a" in terms of m. You need to plug those into a suitable suvat equation and rearrange to make mm the subject.
Original post by old_engineer
No, that's not what I got. Probably best if you post your working.
i cant it says smth goes wrong, would it work if i type it down?
mg-T=ma
mg-T=m((m-1)g/m+1)
mg-T=mg
T=2mg
mg-T=m((m-1)g/m+1)
mg-T=mg
T=2mg
Original post by ugauafidufd
mg-T=ma
mg-T=m((m-1)g/m+1)
mg-T=mg
T=2mg
mg-T=m((m-1)g/m+1)
mg-T=mg
T=2mg
Between your second and third lines of working, (m-1) / (m+1) has disappeared. These terms don't cancel.
Original post by old_engineer
Between your second and third lines of working, (m-1) / (m+1) has disappeared. These terms don't cancel.
would i lose all the marks on the question/part?
Original post by ugauafidufd
would i lose all the marks on the question/part?
Yes you would lose marks. Try again but don't cancel the (m - 1) / (m + 1). I would suggest starting at your second line of working and rearrange to give T = mg - m(m - 1)g / (m+1). Then put the entire right hand side over the denominator (m + 1). Then post your working.
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