Physics 220 class exercise October 18 1999

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:
We can find the tension by using the free-body diagram for the whole system to get the acceleration, and then using the free-body diagram for object 1.  Then we can find the work done on each object individually.  Try it yourself!