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.
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