Question

A Red Rider bb gun uses the energy in a compressed spring to provide the kinetic energy for propelling a small pellet of mass 0.27 grams. By cocking the lever attached to the stock of the rifle, the spring (k = 1800 N/m) is compressed 8.5 cm from its equilibrium point.

Required:
a. What is the potential energy energy stored in the compressed spring?
b. What is the kinetic energy of the pellet when the spring is released?

Answer 1

a. The potential energy stored in the compressed spring is 6.8925 J. b. The kinetic energy of the pellet when the spring is released is 6.8925 J.

Step-by-step explanation:

a. The potential energy stored in the compressed spring can be calculated using the formula U = 1/2kx², where U is the potential energy, k is the spring constant, and x is the displacement from the equilibrium point. In this case, the displacement is given as 8.5 cm (or 0.085 m) and the spring constant is 1800 N/m. Plugging this into the formula, we get U = 1/2 * 1800 N/m * (0.085 m)² = 6.8925 J.

b. The kinetic energy of the pellet when the spring is released can be found using the equation K = 1/2mv², where K is the kinetic energy, m is the mass of the pellet, and v is the velocity. Since the pellet is being propelled by the compressed spring, the potential energy stored in the spring is converted entirely into kinetic energy. Therefore, the kinetic energy is equal to the potential energy, which we calculated in part a to be 6.8925 J.

Answer 2

a.6.5025 J

b.6.5025 J

Step-by-step explanation:

We are given that

Mass of pellet,m=0.27 g=
`0.27* 10^(-3) kg`

1 kg=1000 g

Spring constant,k=1800 N/m

x=8.5 cm=
`8.5* 10^(-2) m`

1m=100 cm

a.Potential energy stored in the compressed spring is given by

P.E=
`(1)/(2)kx^2`

`P.E=(1)/(2)(1800)(8.5* 10^(-2))^2`

`P.E=6.5025 J`

b.By using law of conservation of energy

P.E of spring=K.E of the pellet

K.E of the pellet=6.5025 J