Final answer:
The speed at which the 2-gram BB leaves the spring BB gun, which has a spring constant of 1000 N/m and is compressed 2 cm, is approximately 14.1 m/s, making the correct answer option c) 14.1 m/s.
Step-by-step explanation:
To determine with what speed the 2-gram BB leaves the BB gun, we can use the conservation of energy principle. The potential energy stored in the compressed spring is equal to the kinetic energy of the BB when it leaves the gun, assuming no energy is lost to friction or other forces.
The potential energy (PE) in the spring when compressed is given by ½ k x², where k is the spring constant and x is the compression distance. Using the given values, k = 1000 N/m and x = 2 cm (or 0.02 meters), the calculation would be ½ (1000 N/m)(0.02 m)².
The kinetic energy (KE) of the BB is given by ½ m v², where m is the mass of the BB and v is its velocity. The BB's mass is given as 2 grams, or 0.002 kilograms. Setting the calculated potential energy equal to the BB's kinetic energy, we solve for v to find the speed of the BB as it leaves the gun.
Performing the calculations, the speed v comes out to be approximately 14.1 m/s, so the answer is option c) 14.1 m/s.