Can someone please help with my question?

Question

A toy cannon uses a spring to project a 5.20-g soft rubber ball. The spring is originally compressed by 5.01 cm and has a force constant of 8.01 N/m. When the cannon is fired, the ball moves 15.9 cm through the horizontal barrel of the cannon, and the barrel exerts a constant friction force of 0.032 6 N on the ball.

(a) With what speed does the projectile leave the barrel of the cannon?

 _____ m/s

(b) At what point does the ball have maximum speed?

______ cm (from its original position)

(c) What is this maximum speed?

_____  m/s

Answer 1

Let

m = 5.2g be the mass of the ball,

x(t) be the distance travelled by the ball at time t,

s = 5.01cm be the initial compression of the spring,

k = 8.01N/m be the force constant of the spring,

b = 15.9cm be the length of the barrel (I am assuming that it is the distance from the ball's initial position

to the muzzle),

F_f = 0.0326N be the constant force of friction,

v be the muzzle velocity (to be calculated).

(a)

The energy of the compressed spring, k s²/2,

is converted into the ball's kinetic energy, m v²/2,

plus the work of the force of friction, F_f b.

That is,

k s²/2 = m v²/2 + F_f b

m v² = k s² - 2 F_f b

v = sqrt((k s² - 2 F_f b)/m)

Substituting actual numbers:

v = sqrt((8.01N/m (5.01cm)² - 2 . 0.0326N . 15.9cm)/5.2g)

= sqrt((201 N cm²/m - 1.037 N cm)/5.2g)

= sqrt((0.0201 Nm - 0.01037 Nm)/0.0052kg)

= sqrt(1.872 m²/s²) = 1.37 m/s

(b)

Before the ball emerges from the muzzle it travels through two phases.

First it is being accelerated by the force of the spring reduced by the force of friction.

This accelerating force is maximum at the beginning, and then gradually reduces to 0.

The rest of its trajectory the ball experience only the force of friction, slowing it down.

Therefore the ball has maximum speed at the point when the accelerating force becomes 0.

That happens at the point x where the force of the spring, k(s - x),

reduced by the force of friction,F_f, becomes 0.

k(s - x) - F_f = 0

x = s - F_f/k

Substituting actual numbers,

x = 5.01cm - 0.0326N/(8.01N/m)

= 5.01cm - 0.0326N/(0.0801N/cm) = 4.61cm

(c)

Let v_m be the maximum speed (to be calculated).

The energy of that maximum speed, m v_m²/2, gets converted into

the energy of the muzzle velocity, m v²/2, plus

the work performed by friction F_f(b - x).

Thus

m v_m²/2 = m v²/2 + F_f(b - x)

v_m = sqrt(v² + 2F_f(b - x)/m)

Substituting actual numbers:

vm = sqrt((1.37 m/s)² + 2 . 0.0326N (15.9cm - 4.61cm)/5.2g)

= sqrt(1.872 m²/s² + 2 . 0.0326N . 0.1129m / 0.0052kg)

= sqrt(1.872 m²/s² + 1.416 m²/s²)

= 1.81 m/s