A toy gun uses a spring to project a 6.4 g soft rubber sphere horizontally. The spring constant is 9.0 N/m, the barrel of the gun is 12.4 cm long, and a constant frictional force of 0.028 N exists between barrel and projectile. With what speed does the projectile leave the barrel if the spring was compressed 5.0 cm for this launch? (Assume the projectile is in contact with the barrel for the full 12.4 cm.)

1.584 m/s.

Explanation:

Using,

Ek= Ws-Wf Equation 1

Where Wf = work done against friction, Ek = kinetic energy of the rubber, Ws = work done by the spring.

1/2mv² = 1/2ke²-F'd Equation 2

Where m = mass of the rubber, v = velocity of the rubber, k = force constant of the spring, e = compression, F' = force of friction, d = distance/ length of contact.

make v the subject of the equation

v = √[2(1/2ke²-F'd)/m] Equation 3

Given: k = 9 N/m. e = 5 cm = 0.05 m, F' = 0.028 N, d = 12.4 cm = 0.124 m, m = 6.4 g = 0.0064 kg

Substitute into equation 3

v = √[2(1/2×9×0.05²-0.026×0.124)/0.0064]

v = √(2[0.01125-0.003224]/0.0064)

v = √2.508125

v = 1.584 m/s.

this means that the temperature of both substances will be equal