e 13 8 184

(a) This works just like e_13_8_183
.
K_{f}+U_{f} = K_{i}+U_{i}
\frac{1}{2}mv_{f}^{2}+mgy_{f}+\frac{1}{2}kx_{f}^{2} = \frac{1}{2}mv_{i}^{2}+mgy_{i}+\frac{1}{2}kx_{i}^{2}
\frac{1}{2}(0)^{2}+mgh+\frac{1}{2}k(0)^{2} = \frac{1}{2}(0)^{2}+mg(0)+\frac{1}{2}kx_{i}^{2}.
so
h=\frac{k}{2mg}x_{i}^{2}=\frac{2080\frac{N}{m}}{2(1.93kg)(9.8\frac{m}{s^{2}})}(0.187m)^{2}=1.92m.
The distance up the incline is from trigonometry:
d=\frac{h}{sin\theta}=\frac{1.92m}{sin(27^{o})}=4.23m.