A toy cannon uses a spring to project a 5.27 g soft rubber ball. The spring is o

Question

A toy cannon uses a spring to project a 5.27 g soft rubber ball. The spring is originally compressed by 4.90 cm and has a force constant of 7.90 N/m. When the cannon is fired, the ball moves 14.4 cm through the horizontal barrel of the cannon, and there is a constant frictional force of 0.0311 N between the barrel and the ball. (a) With what speed does the projectile leave the barrel of the cannon? (b) At what point does the ball have maximum speed? from its original position) (c) What is this maximum speed? I did get part A to be 1.38m/s but i cant figure out B or C

Explanation / Answer

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. Of course, 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, which is roughly the answer in the back of the book. They seem to have included an extra significant digit for some reason.

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