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 m_{k}
= 1/Ö 3.
Look at the free body diagrams:
Put a coordinate system with xaxis down the slope and yaxis perpendicular.
Then the displacement is in the xdirection. Let's look at the components
of the forces:
Force

xcomponent

ycomponent

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