Energy - Dynamics and Relativity - page 59

Solution 3.1.2 Recall that the initial velocity is u . The Work-Energy Theorem allows
us to compute the answer straight away:

1
2 m(v 2

u 2 ) − mgy = −

so

u 2

= −

v 2 gy.

Note that we could have just as easily derived this result using Eq. (1.20).

A

Forces of
Constraint

Velocity

h

B

Figure 3.5 A roller-coaster travels along a track. The forces of constraint do no work.

It often happens that a body is forced to follow a well determined path in a
field of force as a result of some constraint. For example, the roller coaster shown
in Figure 3.5 is constrained to follow the tracks. The car begins its journey at A
then descends along the track, rises again before falling to B . If we ignore friction
and air-resistance then the only forces acting on the car are gravity and the normal
forces between the rails and the wheels that serve to keep the roller-coaster on the
track. Because the wheels of a roller-coaster form an interlocking structure with
the track, the normal forces may act either inwardly or outwardly depending on
whether the tendency is for the car to 'push into the track' or 'fly off the rails'. At
some point on the path between A and B we determine the infinitesimal work to be

d W = F · d r = ( − m g + N ) · v d t, (3.10)

where N is the normal force. Since N and v are orthogonal this gives

d W =− m g · v d t =− mg d y. (3.11)

The work done in going from A to B is then obtained by integration:

B y B

W AB = d W =− mg d y = mg(y A − y B ) = mgh. (3.12)

A y A