Asked by Cheese
                Two masses, m1 = 25.0 kg and m2 = 45.0 kg, are attached by a massless rope over a massless pulley. The angle of the frictionless incline is α = 27.00. Assume that m2 starts at rest and falls a vertical distance of 3.00 m. Assume m1 also starts at rest and the rope remains taut. 
a) Determine the total kinetic energy of the two-mass system after m2 has fallen a vertical distance of 3.00 m.
b) Determine how much work the tension in the rope does on m1 as it moves up the incline.
            
        a) Determine the total kinetic energy of the two-mass system after m2 has fallen a vertical distance of 3.00 m.
b) Determine how much work the tension in the rope does on m1 as it moves up the incline.
Answers
                    Answered by
            bobpursley
            
    I assume m1 is on the ramp, going upward.
force of gravity working on m1 down the ramp: m1*g*Sin27
net force then pulling the masses:
m2*g-m1*g*sin27
net force=total mass*a
total mass is m1+m2
solve for a.
then st distance 3.0
Vf^2=2a*3.00 solve for Vf
Total KE= 1/2 (total mass)Vf
Work done by tension
Force*distance
you know distance, force of tension is
m1*g+m2*g*sin27
    
force of gravity working on m1 down the ramp: m1*g*Sin27
net force then pulling the masses:
m2*g-m1*g*sin27
net force=total mass*a
total mass is m1+m2
solve for a.
then st distance 3.0
Vf^2=2a*3.00 solve for Vf
Total KE= 1/2 (total mass)Vf
Work done by tension
Force*distance
you know distance, force of tension is
m1*g+m2*g*sin27
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