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 4.99 cm and has a force constant of 8.09 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 2 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

mass of rubber ball = 5.20 g = 0.0052 kg
x = 4.99 cm = 0.0499 m
k = 8.09 N/m

d = 15.9 cm = 0.159m
Ff = 0.032 N

A)
From Work Energy Theoam we have :

-F.d = total Mechanical Energy
-F.d = 0.5*(m*v^2 - k*x^2) -----------------------------(1)
-0.032 * 0.159 = 0.5*(0.0052*v^2 - 8.09*0.0499^2)
v = 1.38 m/s

B)
at distance x spring is compresed by (0.0499 - x),so
From Work Energy Theoram we have
-Fx = 0.5 * mb*vb^2 + 0.5 *k(0.0499 - x)^2 - 0.5 * k*d^2 ------- (2)

for maximum velocity we ave to diffrentiate above equation :
-f = 0 + k(0.0499 - x)(-1)
0.0499 - x = f/k
x = 0.0499 - f/k
x = 0.0499 - (0.032/8.09)
x = 0.0459 m

C)
Eqn 2 can be written as :

-F*x = (0.5*mb*vb^2) + (0.5*k(0.0499 - x)^2) - (0.5*k*d^2)

-0.032*0.0459 = 0.5*0.0052*vb^2 + 0.5*8.09*(0.0499 - 0.0459)^2 - 0.5*8.09*0.159^2

vb = 6.22 m/s

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