Answer to Question #121231 in Physics for terry

Question

A 2 kg block is attached to a spring (k = 10 N cm-1
) at point
A as shown in the figure. The spring is compressed 50 cm
by the block. When released the block travels over the
frictionless track until it is launched into the air at point B.
Determine the velocity at point B if height, h, is 4.5 m.

Answer 1

The potential energy of the spring will transform into kinetic energy of the block. This kinetic energy, since we ignore friction and air resistance, will be spent to make the block move up the incline. The rest of the kinetic energy will give us the velocity at point B:

$$\frac{kx^2}{2}=\frac{mv^2}{2},\\\space\\ \frac{mv^2}{2}=mgh+\frac{mu^2}{2},\\\space\\ \frac{kx^2}{2}=mgh+\frac{mu^2}{2},\\\space\\ u=\sqrt{\frac{kx^2}{m}-2gh},\\\space\\ u=\sqrt{\frac{1000\cdot0.5^2}{2}-2\cdot9.8\cdot4.5}=6.1\text{ m/s}.$$