Confused! PHysics problem-Changes in Mechanical Energy for NOnconservative forces?

Question

A toy cannon uses a spring to project a 5.30-g soft rubber ball. The spring is originally compressed by 5.00 cm and has a force constant of 8.00 N/m. When the cannon is fired, the ball moves 15.0 cm through the horizontal barrel of the cannon, and there is a constant friction force of 0.032 N b/t the barrel and the ball. (a) With what speed does the projectile leave the barrel of the cannon? (b) At what point does the ball have max speed? (c) What is this max speed?

Answer 1

A nonconservative force is simply a force that changes the total amount of mechanical energy in a system. Friction is the only nonconservative force here. Energy must be conserved in this problem, so the following expression must be true.

ME0 = MEf + Wnc

Where ME0 is the initial mechanical energy, MEf is the final mechanical energy, and Wnc is the work done by nonconservative forces.

We are dealing with spring potential energy and translational kinetic energy, so:

ME = 1/2 k x^2 + 1/2 m v^2

Our work done by friction is the force of friction multiplied by the distance it travels through the barrel.

Wnc = F d

This makes our conservation equation:

1/2 k x0^2 + 1/2 m v0^2 = 1/2 k xf^2 + 1/2 m vf^2 + F d

We know that xf and v0 both are zero so the equation simplifies to:

1/2 k x0^2 = 1/2 m vf^2 + F d

You just solve for vf and plug your numbers in and calculate for part a.

Answer 2

The spring has an vigor Es = a million/two kx^two earlier than the cannon shoots. The drive of friction does paintings Wf = Ff*d in which d = 15 cm and Ff is the consistent friction drive. So on the finish of the barrel, the vigor final is Es - Ws = a million/2mv^two. Now, does the 15 cm comprise the five cm that the spring is compressed? If no then max pace => a million/2kx^two = a million/2mv^two so vmax = sqrt(ok/m)*x in which x = five cm. If sure, then the calculation is extra elaborate so I suspect the above reply is what's being sought.