The potential energy stored in the compressed spring, when the spring compression distance is X = 0.05 m, is
Ep = (1/2) k X^2 = (0.5)*8.0*(0.05)^2
= 0.0100 Joules
Make sure that the units of the spring constant are really N/m and not N/cm. That is a very small amount of stored spring energy.
The kinetic energy of the sphere leaving the gun equals the compressed spring potential energy MINUS the work done against friction. The friction work is 0.032 N * 0.15 m = 0.0048 J
That leaves
Ek = 0.0052 J for kinetic energy
0.0052 = (1/2) M V^2
Mass M = 0 0053 kg , so
V^2 = 2*.0052/0.0053 = 2
V = 1.4 m/s
A toy gun uses a spring to project a 5.3 g soft rubber sphere vertically. The spring constant is 8.0 N/m, the barrel of the gun is 15 cm long, and a constant frictional force of 0.032 N exists between barrel and projectile. How high does the projectile go if the spring was compressed 5.0 cm for this launch? For this, I need the problem solved step by step.
1 answer
1 answer