Help me please?

Question

Two masses, m1 = 28.0 kg and m2 = 42.0 kg, are connected by a rope that hangs over a pulley. The pulley is a uniform cylinder of radius 0.50 m and mass 5 kg. Initially m1 is on the ground and m2 rests 2.5 m above the ground. If the system is released, use conservation of energy to determine the speed of m2 just before it strikes the ground. Assume the pulley bearing is frictionless.
m/s

Answer 1

Before anything moves the system's potential energy is just the gravitational potential energy of the bigger mass:
PE1 = 42*2.5*9.8 = 1029.69825 J .....................(1)
Right before the big mass hits the ground the energy of the system is the gravitational potential energy of the smaller mass plus the rotational energy of the pulley plus the kinetic energy of the masses:
PE2+KEp+KEm = 28*2.5*9.8 + .5*(.5*5*.5^2)*(v/.5)^2 + .5*(42+28)*v^2 = 686.4655 J + (36.25kg)*v^2 .....................(2)
Now set the expression (1) equal to (2) since the energy before is equal to the energy after, and solve this for v:
(1029.69825 J - 686.4655 J) / (36.25kg) = v^2
and you get v = 3.1 m/s.

Answer 2

Is this an assignment ? I don't want to give you the exact answer, but here is a hint, use the formula for force (F = ma), use it to equate both sides of the system. If I'm not mistaken, the size & massof the pulley is not significant, it was only added to confuse you.

Answer 3

That's a tough situation. To be honest, I never dealt with anything like that in my life, so i really can't see from your perspective. However, you could tell your dad how you feel about living with your grandma, and maybe he will do something. I am sure that at least one of your parents would listen to your problem and take action, so try it out and see what they have to say about it.

Answer 4

You will need a few equations
1. I= .5m r^2 (moment of inertia for the pulley)
2. Er= .5Iw^2 (rtational energy of the pulley)
3. Pe= mgh (potential energy)
4. Pk= .5mv^2 (kinetic enrgy)
5. w=2 Pi v/ r

System equation
Pe(total)=Pk(total)
Pe(total)=Pe2-Pe1= (m2-m1)gh

Pk(total)=Pk1+Pk2+Er(of the pulley)
Pk(total) = .5(m2+m1)V^2 + .5Iw^2
Pk(total) =.5(m2+m1)V^2 + .5(.5m r^2 )w^2

(m2 - m1)gh=.5(m2 + m1)v^2 +.25(m r^2 )w^2
(m2 - m1)gh=.5(m2 + m1)v^2 +.25(m r^2 )( 2Pi V/r)^2
(m2 - m1)gh=.5(m2 + m1)v^2 + m( Pi v)^2

If my algebra is correct;

V=sqrt((m2-m1)gh/((m2+m1)/2 + Pi^2 m/)

Use m1= 28kg; m2=42kg; m= 5kg; r =.5m
H=2.5m

Just use a calculator.
1. (m2-m1)g h=343 joules
2. (m2+m1)/2 = 35 kg
3. Pi^2 m=49kg

V=sqrt(343/(35+49))= 2 m/s

Answer 5

Do ur Home work!!!!!!!!!