Energy and WET (Springs)

In summary, a toy cannon uses a spring with a force constant of 9.00 N/m and initially compressed by 5.00 cm to project a 5.30 g rubber ball through a 16.0 cm horizontal barrel. The frictional force between the barrel and ball is 0.0320 N. To find the speed of the ball, we use the equation Wnet = \DeltaKE, where Wnet is the net work done on the ball, \DeltaKE is the change in kinetic energy, and Ws and Wfk are the spring work and friction work, respectively. By substituting the given values and solving for v, we get a final speed of 1.7079 m/s. However
  • #1
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A toy cannon uses a spring to project a 5.30 g soft rubber ball. The spring is originally compressed by 5.00 cm and has a force constant of 9.00 N/m. When it is fired, the ball moves 16.0 cm through the horizontal barrel of the cannon, and a constant frictional force of 0.0320 N exists between barrel and ball.

(a) With what speed does the projectile leave the barrel of the cannon?

Wnet = [tex]\Delta[/tex]KE
Ws - Wfk = .5mv2 - 0
.5kx2 - fkd = .5mv2
.5(9)(.052) - (.032)(.11) = .5(.0053)v2
v = 1.7079

Final answer is incorrect. Any ideas?
 
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  • #2
Have you tried replacing .11 with .16? Friction acts over full length of barrel. What is the answer?
 
  • #3


There may be some errors in the calculations. Let's break down the steps to find the speed of the projectile leaving the barrel:

1. First, we need to find the total work done on the ball by the spring. This can be calculated using the formula Ws = 0.5kx2, where k is the force constant and x is the compression distance.

Ws = 0.5(9 N/m)(0.05 m)^2 = 0.01125 J

2. Next, we need to calculate the work done by friction. This can be calculated using the formula Wfk = fkd, where f is the frictional force and d is the distance the ball moves.

Wfk = (0.032 N)(0.16 m) = 0.00512 J

3. Now, we can use the work-energy theorem to find the change in kinetic energy of the ball:

Wnet = ΔKE
0.01125 J - 0.00512 J = 0.5mv2 - 0

4. Rearranging the equation, we get:

0.5mv2 = 0.00613 J

5. Finally, we can solve for the velocity of the ball using the mass and the calculated work:

v = √(2W/m) = √(2(0.00613 J)/0.0053 kg) = 1.76 m/s

Therefore, the speed of the projectile leaving the barrel is approximately 1.76 m/s. It is important to double check the calculations and make sure the units are consistent throughout the problem.
 

1. What is energy?

Energy is the ability to do work or cause change. It is a fundamental concept in physics and is measured in joules (J).

2. How is energy related to springs?

Springs store potential energy when they are stretched or compressed. This potential energy is converted into kinetic energy as the spring returns to its original shape.

3. What factors affect the amount of energy stored in a spring?

The amount of energy stored in a spring depends on its stiffness or spring constant (k) and the displacement from its equilibrium position (x). The formula for potential energy stored in a spring is U = 1/2 * k * x^2.

4. How does energy transfer in a spring system?

In a simple spring system, energy is transferred back and forth between potential energy and kinetic energy as the spring oscillates. In more complex systems, energy can also be lost through friction or air resistance.

5. What is the difference between elastic and inelastic collisions in terms of energy and springs?

In an elastic collision, the total kinetic energy is conserved as the objects bounce off each other. In an inelastic collision, some of the kinetic energy is lost and converted into other forms of energy, such as heat or sound. In terms of springs, an elastic collision would result in a spring returning to its original shape, while an inelastic collision may cause permanent deformation of the spring.

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