please help me with this hard physics problem?

Question

masses m1= 4 kg and m2=9 kg are connected by a light string that passes over a frictionless pulley. m1 is held at rest on the floor and m2 rests on a fixed incline of theta is 40 degrees. the masses are released from rest and m2 slides 1 meter down the incline in 4 seconds. determine the acceleration of each mass, the tension in the string and the coefficient of kinetic friction between m2 and the incline

Answer 1

Where is the pulley? I will assume the top of the pulley is exactly even with the plane of the incline so that as mass m2 moves down the incline 1 m, mass m1 rises 1 m. Also, I will assume the 40 degree incline is 40 degrees w/r/t horizontal.

Consider a fbd of m1, with positive in the direction of motion. T is the tension in the string
T-4*g=4*a
T=4*(g+a)

Now consider a fbd of m2 evaluating forces parallel to the surface of the incline
9*g*sin(40)-T-9*g*µk=9*a

T=9*(g*(sin(40)-µk)-a)

From the givens, we can calculate a
s(t)=.5*a*t^2
s(4)=1
1=.5*a*4^2
solve for a
a=0.125 m/s^2
Plug this into the first equation from the m1 fbd above
T=4*(g+0.125)
solve for T
T=39.74 N
Plug this into the second equation along with a
and
39.74=9*(g*(sin(40)-µk)-0.125)
Solve for µk
µk=0.18

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Answer 2

Using v = u + at in which u is preliminary pace and v last pace we all know that during each circumstances v = zero on the grounds that each autos come to relaxation. Initial pace u: u + (-four.two)t1 = zero or u = four.2t1 Initial pace 2u : making use of identical inspiration 2u = four.2t2 Dividing moment equation via first 2u / u = four.2t2 / four.2t1 t2 / t1 = two t2 = two t1 This indicates the moment automobile takes two times the time to quit. u1 = sixteen m/s so t1 = sixteen / four.two = three.eighty one s u2 = 32 m/s so t2 = 32 / four.two = 7.sixty two s

Answer 3

Do you have an actual question to ask about this problem, or were you planning on having others do your entire homework set for you?