Masses, Pulley, Incline, Friction.. PLease Help..All geniuses out there!! PLease be detailed, i am so confused

Question

A 12.0 kg block (m1) is on an 29degree incline. It is attached to a 10.0kg block (m2) by a massless cable streched over a massless, frictionless pulley. Both blocks are at a height of 1.5m when released.
Assume the ramp has a kinetic friction coefficient of 0.320
a) What is the speed of the masses just before m2 hits the ground?
b)What is the total Kinetic Energy of the system just before m2 hits the ground?
c)Find the total mechanical energy just before m2 hits the ground.
d)How much energy is lost due to friction?

Answer 1

The force pulling block1 up the ramp is m2*g. It is opposed by the component of weight of the block1 parallel to the ramp and the frictional force from the ramp. The weight component is m1*g*sinø (ø is the angle of the ramp). The frictional component is k*m1*g*cosø, where k is the coeff of friction. Therefore the total force acting on m1 is:

F = m2*g - m1*g*sinø - k*m1*g*cosø

The acceleration of the whole system is a = F/m1, or

a = [(m2/m1)* - sinø - k*cosø]*g

The distance an object travels under constant acceleration a is
s =.5*a*t^2 solve for t: t = 2√[s/a]. The velocity after time t will be a*t, so v = 2√[as].

Both masses are moving at the same speed.

Kinetic energy is .5*m*v^2, where m is the total mass of the system (m1 + m2) at velocity v.

The total energy will be the kinetic energy above plus the gain in potential energy of m1. That is m1*g*h, where h is the altitude gained by m1. m1 went up 1.5m on a slope of øº, so sinø = 1.5/h, solve for h.

Energy lost to friction is Fr*s, where Fr is the frictional force and s is the distance traveled. You have a the formula for frictional force above, and the distance traveled of 1.5m

Answer 2

♠ thus when they are both at rest the initial pot energy of the system is W=m2*g*h, where h=1.5m;
♣ when m2 reaches the floor the kin energy of the system E=E1+E2, where
E1=0.5*m1*v^2, E2=0.5*m2*v^2, where v is their common speed for they are attached to the same cable;
♦ now energy conservation law: initial pot energy W=E1+E2+W1+Wf, where W1 is work done by m2 to lift m1 up the slope 29°, that is wok against gravity W1=m1*g*h1, where h1=h*sin29°; work of friction Wf = u*Ff*h, where u*Ff is force of friction, Ff=m1*g*cos29° is component of weight m1*g normal to the slope 29°, u=0.32, h=1.5m is the pass on which friction occurs;
thence plugging numbers, d) Wf= u*m1*g*cos29*h =49.37 J;
also W-W1-Wf=E; or;
(m2 -m1*sin29 -u*m1*cos29)*gh =0.5(m1+m2)*v^2, hence
v^2=2*(m2- m1*(sin29°+u*cos29))*gh / (m1+m2);
♥ now numbers come:
a) v^2= 2*(10-12*(sin29+0.32*cos29)) *9.8*1.5 /(12+10) =
= 1.100823; v=1.0492m/s;
b) E= 0.5(10+12)* 1.100823= 12.11 Joules;
c) since some energy was wasted on friction then
W-Wf =10*9.8*1.5 –49.37 = 97.63 J;

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