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!