56. To solve the problem, we note that the acceleration along the slanted path depends on only the force components along the path, not the components perpendicular to the path. (a) From the free-body diagram shown, we see that the net force on the putting shot along the +x-axis is Fnet, x = F − mg sin θ = 380.0 N − (7.260 kg)(9.80 m/s 2 ) sin 30° = 344.4 N,
which in turn gives ax = Fnet, x / m = (344.4 N) /(7.260 kg) = 47.44 m/s 2 .
Using Eq. 2-16 for constant-acceleration motion, the speed of the shot at the end of the acceleration phase is
v = v02 + 2ax Δx = (2.500 m/s) 2 + 2(47.44 m/s 2 )(1.650 m) = 12.76 m/s. (b) If θ = 42°, then ax =
Fnet, x m
=
F − mg sin θ 380.0 N − (7.260 kg)(9.80 m/s 2 ) sin 42.00° = = 45.78 m/s 2 , m 7.260 kg
and the final (launch) speed is v = v02 + 2ax Δx = (2.500 m/s) 2 + 2(45.78 m/s 2 )(1.650 m) = 12.54 m/s.
(c) The decrease in launch speed when changing the angle from 30.00° to 42.00° is 12.76 m/s − 12.54 m/s = 0.0169 = 16.9%. 12.76 m/s
