OCR (MEI) Mechanics 1 -Connected Particles

I can't figure out how part i) of this question works and the book wasn't much use either. I have given the solutions form the markscheme. part ii) i do not need help with, but i decided it anyway though.
Can someone please give me a good explanation of how to get the equations of motion. cheers

Question 5 Wednesday 10th January 2007

look at the attached diagram

A 4kg box rests on a horizontal plane with constant resistance of 14.8N. A weightless inextensile string is attached the block and after passing over the resistanceless pulley is attached a 2kg mass. the string between the box and pulley is horizontal

i) Write down an equation of motion for the block and also the equation of motion of the 2kg mass in terms of T and a

SOLUTION

4kg Box: T-14.8= 4a
2kg Mass: 2g-T= 2a

ii) find the values of T and a

SOLUTION

4kg Box: T-14.8= 4a
2kg Mass: 2g-T= 2a

add them and rearrange to get;

a = (2g-14.8)/6 = 0.8

therefore

T= 4a + 14.8
T= 3.2 + 14.8
T= 18n

Mechanics.bmp113.9KB

Reply 1

generalebriety

Newton's second law (F=ma).

Reply 2

gam3r

I just did that exam as a mock, i found it rather difficult compared to the other past papers i did. I also got confused on this question and i couldn't do the question with the slope in it either, i just don't get it..

Reply 3

Lucien_Roach

hmmm had another shot at it

gonna say D is delta cause i'm lazy

DF=ma
m=m1+m2 , R = resistance

for Force on 4kg Block (m2).

two forces acting T and R

F = T - R = m2a

T -14.8 = 4a

For Force on 2kg block (m1).

again two forces weight and tension (opposite direction)

M1g-T=2a