A toy cannon uses a spring to project a 5.30-g scott rubber ball. The spring has

Question

A toy cannon uses a spring to project a 5.30-g scott rubber ball. The spring has a stiffness constnat of 8.00 N/m and is originally compressed by 5.00 cm to point A in the figure. When the cannon is fired, the ball moves a total of 15.0 cm through the barrel of the cannon (from point A to point C), and there is a constant frictional force of 0.0320 N between the barrel and the ball (from point A to point C as well). (The spring move past B, its equilibrium point.) How much work is done by the spring as it moves from its compressed point and returns to equilibrium? How fast is the ball moving when it reaches the equilibrium point of the spring? How fast is the ball moving when it leaves the barrel of the cannon? If the cannon is pointed straight upwards, how high will the ball travel? Assume that the ball leaves the cannon with the speed you calculated in part c.

Explanation / Answer

a) W = 0.5kx^2 = 0.5*8*(0.05)^2 = 0.01 J
b) Energy lost due to friction = 0.032*0.05 = 0.0016 J
Net work done on the ball = 0.0084 J = 0.5mv^2
v = 1.78 m/s
c) Net energy lost = 0.032*0.2 = 0.0064 J
0.5mv^2 = 0.01 - 0.0064
v = 1.16 m/s
d) 2*9.8*h = 1.16^2
h = 0.0693 m

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