Question

A spring with spring constant 40 N/m is compressed 0.1 m past its natural

length. A mass of 0.5 kg is attached to the spring.
a. What is the elastic potential energy stored in the spring? (1 point)
b. The spring is released. What is the speed of the mass as it reaches the natural
ength of the spring? (1 point)

Answer 1

a). The elastic potential energy stored in the spring is 0.2 J.

b). The speed of the mass when it reaches the natural length of the spring is 0.2 m/s.

Step-by-step explanation:

a. To calculate the elastic potential energy stored in the spring, we can use the formula:
Elastic potential energy = 0.5 * k * x^2
where k is the spring constant (40 N/m) and x is the displacement (0.1 m).
Plugging in the values, we get:
Elastic potential energy = 0.5 * 40 * (0.1)^2
Elastic potential energy = 0.2 J

b. When the spring is released, the elastic potential energy is converted into kinetic energy.

We can use the law of conservation of energy to find the speed of the mass.
Initial potential energy = Final kinetic energy
0.5 * k * x^2 = 0.5 * m * v^2
Plugging in the values, we get:
0.5 * 40 * (0.1)^2 = 0.5 * 0.5 * v^2
Solving for v, we get:
v = sqrt([(0.5 * 40 * (0.1)^2) / (0.5 * 0.5)])
v = 0.2 m/s