MHB Work energy principle and power

[Shah 72]

Particle X of mass 2 kg , and particle Y of mass m kg are attached to the ends of a light inextensible string of length 4.8m. The string passes over a fixed smooth pulley and hangs vertically either side of the pulley. Particle X is held at ground level, 3m below the pulley. Particle X is released and rises while particle Y descends to the ground

a) Find an expression in terms of m for the tension in the string while both particles are moving.
By getting two equations
T-20=2a and T-10m=-ma
Solving simultaneously and removing a I got mT+2T-40m=0,
I finally got T=40m/(m+2) N
b) use work energy principle to find how close particle X gets to the pulley in subsequent motion.
Iam not able to get this ans. Pls help

 

[skeeter]

work done on the 2kg mass by the force of tension ...

$W = (T \, Newtons) \cdot (1.2 \, meters) = 2gH$, where $H$ is the highest point mass X rises above ground level.

 

[Shah 72]

work done on the 2kg mass by the force of tension ...

$W = (T \, Newtons) \cdot (1.2 \, meters) = 2gH$, where $H$ is the highest point mass X rises above ground level.

Thanks!

 

[Kansas Boy]

The problem I have with
"a) Find an expression in terms of m for the tension in the string while both particles are moving.
By getting two equations
T-20=2a and T-10m=-ma"
is that you have not said what either "T" nor "a" are!

I can guess that "T" is the tension in the string and that "a" is the acceleration of the particles but you really should have said thar.