Physics...energy and work?!?

Question

A small block of mass 0.09 kg is placed against a compressed spring at the bottom of a frictionless track that slopes upward at an angle of 40 degrees above the horizontal. The spring has k = 640 N/m and negligible mass. When the spring is released, the block travels a max distance of 1.8 m along the track before sliding back down. Before reaching this max distance, the block loses contact with the spring. (a) what distance was the spring originally compressed? (b) When the block has traveled along the track 0.8 m from its initial position against the compressed spring, is it still in contact with the spring? What is the kinetic energy of the block at this point?

Answer 1

Hi
The vertical height reached by the block is 1.8sin40°=1.157m.
Its mass is 0.09g so the total energy at its highest point is Ep=mgh=0.09*9.8*1.157=1.0205 J.
Originally, this energy came from the compressed spring's elastic potential energy, the formula for which is ½kx².

½kx²=1.0205
x²=2*1.0205/640
x=0.0565m

After travelling 0.8m the block will not still be in touch with the spring. Even an ideal spring will only extend the same amount it was compressed. It will have gained a height of 0.8sin40=0.5142m so its gravitational potential will be 0.09*9.8*0.5142=0.4536J.
subtract this from the total energy and the "left-overs" must be kinetic energy.

Answer 2

The block gained potential energy in the amount of m*h, where h is the altitude gained. h = L*sinø, where L = distance traveled along the ramp, and ø the angle of the ramp from the horizontal.

This much energy must be supplied by the energy stored in the spring, which is 0.5*k*∆s^2, where k=spring constant and ∆s the compression of the spring.

Equating these gives m*g*L*sinø = 0.5*k*∆s^2; solve for ∆s

∆s = √[2*m*g*L*sinø / k]

∆s = 0.056 m

Since this is less than 0.8 m, the block has left the spring.

The kinetic energy is the energy from the spring less the potential energy at 0.8 m

0.5*k*∆s^2 - m*g*0.8*sin(40) = 0.56 J

The velocity is given by 0.56 = 0.5*m*v^2

v = 3.52 m/sec