6.2. WORKED EXAMPLES
System for Example 16
v = 0
After sliding down the slope and going over the rough part, the mass has maximally squished
the spring by an amount x = 0.300 m.
this case we can calculate the work that it does. Then, we can use the energy conservation
K + U = W
to find the unknown quantity in this problem, namely µ
for the rough surface. We can
get the answer from this equation because we have numbers for all the quantities except for
which depends on the coefficient of friction.
The block is released at point A so its initial speed (and hence, kinetic energy) is zero:
= 0. If we measure height upwards from the level part of the track, then the initial
potential energy for the mass (all of it gravitational) is
= mgh = (10.0 kg)(9.80
)(3.00 m) = 2.94 × 10
Next, for the "final" position of the mass, consider the time at which it has maximally
compressed the spring and it is (instantaneously) at rest. (This is shown in Fig. 6.17.) We
don't need to think about what the mass was doing in between these two points; we don't
care about the speed of the mass during its slide.
At this final point, the mass is again at rest, so its kinetic energy is zero: K
= 0. Being
at zero height, it has no gravitational potential energy but now since there is a compressed