Physics 220 class exercise October 20 2000

Work done on a system of particles.
 
 

The system consists of two objects sliding down a ramp making an angle of 30° with the horizontal . Friction acts only on the second object of the system, and mk = 1/Ö 3.

Look at the free body diagrams:

Put a coordinate system with x-axis down the slope and y-axis perpendicular. Then the displacement is in the x-direction. Let's look at the components of the forces:
Force x-component y-component
n1 0 n1
T1 - T1 0
m1g m1gsinq - m1gcosq
n2 0 n2
T2 T2 0
f -f 0
m2g m2gsinq - m2gcosq

  Now both objects accelerate together down the slope, and the acceleration has no y component. So for object two we may use Newton's 2nd law to conclude:
  å Fy = n2 - m2gcosq = 0
 
 
and thus n2 = m2gcosq . Then the kinetic frictional force is f = mk m2gcosq .
    Now we are ready to find the work done. For the system as a whole we may use the diagram on the far right. The normal forces do no work since they are perpendicular to the displacement. The work done is thus:  

By the work energy theorem, the work done equals the change in kinetic energy of the system:

 
W = Kf - KI = (m1+m2)v2/2.
 


Thus the final speed of the system is:

 
Now we can put in some numbers:
  To find the work done on each object individually, first find the tension.  Begin with the free body diagram for the whole system, and use Newton's second law to find the acceleration.  The go to the free-body diagram for object 1, use your known acceleration to find T.  Then you can find the work.  Try it yourself.