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

(b)
Because friction applies a constant backward (let's assume leftward) force, the ball will only accelerate to the right if the spring force exceeds the force of friction. So let's set the two equal to see when this ceases to happen.
F_spring = F_friction
kx = 0.032 N
(8.00 N/m)(x) = 0.032 N
And we find that x = 0.0040 m or 0.40 cm. Of course, this is how much the spring is compressed. Since the spring originally was compressed by 5.00 cm, the ball has now traveled a distance of 4.60 cm.

(c)
First, let's consider how much work is done on the ball as it travels this distance by the two forces we need to consider: namely, the spring force and friction. We can use W = 1/2 k x^2 to calculate the spring's work, but keep in mind that the spring hasn't finished releasing, so it will actually look like this:
W_spring = 1/2 * ( 8.00 N/m) * [(0.0500 m)^2 - (0.0040 m)^2]
W_spring = 0.0099 J

The work done by friction is easy because it is a constant force, so
W_fric = Fd
W_fric = (-0.032 N) (0.0460 m)
W_fric = -.0015 J

I have called the frictional force negative because it is pushing the ball left, and since it is opposite the displacement it makes negative work. So adding the works, we get 0.0084 J. , we need to know the velocity after this net work has been applied, so we use the work-energy theorem to find that
K = W (since K_initial was zero)
1/2 m v^2 = 0.0084 J
1/2 (0.00530 kg) (v^2) = 0.0084 J
And solving for v, we get v = 1.8 m/s

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