*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 m_{k}
= 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 |

n_{1} |
0 |
n_{1} |

T_{1} |
- T_{1} |
0 |

m_{1}g |
m_{1}gsinq |
- m_{1}gcosq |

n_{2} |
0 |
n_{2} |

T_{2} |
T_{2} |
0 |

f |
-f |
0 |

m_{2}g |
m_{2}gsinq |
- m_{2}gcosq |

Now both objects accelerate together down the slope, and the acceleration
has no y component. So for object two we may use Newton's 2^{nd}
law to conclude:

å F_{y} = n_{2 }-
m_{2}gcosq = 0

and thus n_{2 }= m_{2}gcosq
. Then the kinetic frictional force is f = m_{k}
m_{2}gcosq .

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 = K_{f} - K_{I} = (*m*_{1}*+m*_{2})v^{2}/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.