A toy cannon uses a spring to project a 5.22-g soft rubber ball. The spring is o

Question

A toy cannon uses a spring to project a 5.22-g soft rubber ball. The spring is originally compressed by 4.96 cm and has a force constant of 8.04 N/m. When the cannon is fired, the ball moves 14.1 cm through the horizontal barrel of the cannon, and the barrel exerts a constant friction force of 0.033 0 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

Explanation / Answer

(a)

The initial potential energy of the spring is:
Ep = k×x²/2 = 8.08×0.0496²/2 = 0.009939 J

The work done by friction is:
W = F×s = 0.033×0.141 = 0.004653 J

So the remaining kinetic energy of the projectile at the end of the barrel is:
Ek = Ep - W = 0.009939 - 0.004653 = 0.005286 J

and its velocity will be:
0.005286 = 0.00522×v²/2

0.010572 = 0.00522 ×v²
v = 1.423 m/s < - - - - - answer (a)

(b)

The maximum speed will be when the spring reaches its position of equilibrium, so
at 4.96 cm (from its original position) < - - - - - answer (b)

(c)

The work done by friction is:
W = F×s = 0.033×0.0496 = 0.001636 J

So the remaining kinetic energy of the projectile at the spring's equilibrium position is:
Ek = Ep - W = 0.009939 - 0.001636 = 0.008303 J

and its velocity will be:
0.008303 = 0.00522×v²/2

0.016606 = 0.00522 ×v²
v = 1.783 m/s < - - - - - answer (c)

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