The spring constant for the spring in a special cannon is 1800 N/m. In cocking the cannon, the spring is compressed 0.55 m. What is the initial speed of a 7.0-kg cannonball at rest on the free end of the spring when it is released?
A)77 m/s
B)140 m/s
C)12 m/s
D)8.8 m/s
E)16 m/s

Question

Quizplus · Accepted Answer

The potential energy stored in the spring when it is compressed is given by: 
U = (1/2) k x^2 
where k is the spring constant and x is the displacement of the spring from its equilibrium position. 
Substituting the given values, we get: 
U = (1/2) * 1800 N/m * (0.55 m)^2 = 271.05 J 
This potential energy is converted into kinetic energy of the cannonball when the spring is released. 
Therefore, we can write: 
U = (1/2) mv^2 
where m is the mass of the cannonball and v is its velocity. 
Rearranging, we get: 
v = sqrt(2U/m) = sqrt(2*k*x^2/m) 
Substituting the given values, we get: 
v = sqrt(2*1800 N/m * (0.55 m)^2 / 7 kg) = 139.9 m/s 
Therefore, the initial speed of the cannonball is approximately 140 m/s.

The Spring Constant for the Spring in a Special Cannon