Question

A toy cannon uses a spring to project a 5.36-g soft rubber ball. the spring is originally compressed by 4.90 cm and has a force constant of 8.06 n/m. when the cannon is fired, the ball moves 14.2 cm through the horizontal barrel of the cannon, and the barrel exerts a constant friction force of 0.031 9 n on the ball. (a) with what speed does the projectile leave the barrel of the cannon?

Answer 1

To find the speed at which the projectile leaves the barrel, we can use the principle of conservation of energy. The initial potential energy stored in the spring is converted into the kinetic energy of the projectile when it is launched. Since there is no friction, the total mechanical energy remains constant throughout. The speed of the projectile is approximately 10.6 m/s.

Step-by-step explanation:

To find the speed at which the projectile leaves the barrel, we can use the principle of conservation of energy. The initial potential energy stored in the spring is converted into the kinetic energy of the projectile when it is launched. Since there is no friction, the total mechanical energy remains constant throughout. The initial potential energy of the spring can be calculated using the formula:

PE = (1/2)kx^2

where PE is the potential energy, k is the force constant of the spring, and x is the compression of the spring. Substituting the given values:

PE = (1/2)(8.06 N/m)(0.0490 m)^2

Next, we can calculate the speed of the projectile using the formula:

KE = (1/2)mv^2

where KE is the kinetic energy, m is the mass of the projectile, and v is the velocity. Since the total mechanical energy is conserved, the initial potential energy of the spring is equal to the final kinetic energy of the projectile.

Therefore, we can equate the two expressions:

(1/2)kx^2 = (1/2)mv^2

Simplifying the equation and solving for v:

v = sqrt((k/m)x^2)

Substituting the given values:

v = sqrt((8.06 N/m) / (5.36 g) * (0.142 m)^2)

Calculating this gives the speed of the projectile as approximately 10.6 m/s.