Problem 6.49

6.49.

In this problem we want to know the velocity of the ball as it leaves the launcher. The initial energy in the spring after the launcher is pulled back is

$U_{0} = \frac{1}{2} k x^{2} = \frac{1}{2}(120)(.05)^{2} J$

U₀ = .150 J

Now, when the ball gets to the launch spot it will have both potential and kinetic energy. The potential energy at that point is given by

$U_{f} = mgh = mgx\sin 10{}^{\circ}$

$U_{f} = (.1)(9.8)(.05)\sin 10{}^{\circ}J ~=~ .00851 J$

The kinetic energy of the ball at the launch spot is given by

$K_{f} = \frac{1}{2} m v_{f}^{2}$

we now use the fact that

U₀ = U_f + K_f

.150 J = .00851 J + (.05 kg) v²

solving for v yields

$v = \sqrt{\frac{.141}{.05}} ~=~ 1.68 m/s$