A toy cannon uses a spring to project a 5.20-g soft rubber ball. The spring is o

Question

A toy cannon uses a spring to project a 5.20-g soft rubber ball. The spring is originally compressed by 4.97 cm and has a force constant of 7.94 N/m. When the cannon is fired, the ball moves 15.5 cm through the horizontal barrel of the cannon, and the barrel exerts a constant friction force of 0.032 1 N on the ball. (a) With what speed does the projectile leave the barrel of the cannon? 1.36 Correct: Your answer is correct. m/s (b) At what point does the ball have maximum speed? Incorrect: Your answer is incorrect. cm (from its original position) (c) What is this maximum speed? m/s

Explanation / Answer

(a) The initial potential energy of the spring = Ep = k×x²/2 = 7.94×0.0497²/2 = 0.009806 J
The work done by friction = W = F×s = 0.0321×0.155 = 0.0049755 J

So the remaining kinetic energy of the projectile at the end of the barrel is:
Ek = Ep - W = 0.009806 - 0.0049755 = 0.0048305 J
and its velocity will be:
0.0048305 = 0.0052×v²/2
v = 1.363 m/s

(b) The maximum speed will be when the spring reaches its position of equilibrium, so at 4.97 cm (from its original position)

(c) The work done by friction = W = F×s = 0.0321×0.0497 = 0.00159537 J

So the remaining kinetic energy of the projectile at the spring's equilibrium position is:
Ek = Ep - W = 0.009806 - 0.00159537 = 0.00821 J

and its velocity will be:
0.00821 = 0.00520×v²/2
v = 1.777 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.