p08_012.pdf

12. We use Eq. 8-17, representing the conservation of mechanical energy (which neglects friction and other

dissipative eﬀects).

(a) In the solution to exercise 4, we found U A = mgh (with the reference position at C ). Referring

again to Fig. 8-26, we see that this is the same as U 0 which implies that K A = K 0 and thus that

v A = v 0 .

(b) In the solution to exercise 4, we also found U B = mgh/ 2. In this case, we have

K 0 + U 0 = K B + U B

2 mv B + mg h

2 mv 0 + mgh =

which leads to v B = v 0 + gh .

(d) To ﬁnd the “ﬁnal” height, we set K f = 0. In this case, we have

K 0 + U 0 = K f + U f

2 mv 0 + mgh =0+ mgh f

which leads to h f = h + v 0 / 2 g .

(e) It is evident that the above results do not depend on mass. Thus, a diﬀerent mass for the coaster

must lead to the same results.