The staples inside a stapler are kept in place by a spring with a relaxed length of 0.115 m. If the spring constant is 51.0 N/m, how much elastic potential energy is stored in the spring when its length is 0.150 m?
E = 0.5*k*x^2 = (0.5)*(51N/m)*(.15m - .115m)^2 = .031238 J
thankyou
The staples inside a stapler are kept in place by a spring with a relaxed length of 11.5 cm. The spring constant is 51.0 N/m. How much elastic potential energy is stored in the spring when its length is 8 cm?
31.2375
Why did the stapler go to therapy? Because it had issues with its spring tension!
To calculate the elastic potential energy stored in the spring, we can use the equation:
Elastic Potential Energy = (1/2)k(x^2)
where k is the spring constant and x is the displacement from the relaxed length.
Given:
Spring constant (k) = 51.0 N/m
Displacement (x) = 0.150 m
Let's calculate the elastic potential energy stored in the spring:
Elastic Potential Energy = (1/2) * 51.0 N/m * (0.150 m - 0.115 m)^2
Elastic Potential Energy = (1/2) * 51.0 N/m * (0.035 m)^2
Elastic Potential Energy ≈ 0.031 N·m (or Joules)
So, the elastic potential energy stored in the spring when its length is 0.150 m is approximately 0.031 Joules.
To calculate the elastic potential energy stored in the spring, we can use the formula:
Elastic Potential Energy = (1/2) * k * Δx^2
Where:
- k is the spring constant
- Δx is the change in length of the spring
In this case, the spring constant (k) is given as 51.0 N/m, and the change in length (Δx) is 0.150 m - 0.115 m = 0.035 m.
Now we can substitute these values into the formula:
Elastic Potential Energy = (1/2) * 51.0 N/m * (0.035 m)^2
Elastic Potential Energy = (1/2) * 51.0 N/m * 0.001225 m^2
Elastic Potential Energy = 0.03124 J
So, the elastic potential energy stored in the spring is approximately 0.03124 J.