what is the spring potential energy of a spring that is stretched 15 cm if its spring constant is 350 n/m?
how much force is required for this stretch?
Correct
Ok so this is what I got for the first part
U= 1/2k(delta x)^2
U= 1/2( 350 N/m)(15 cm)^2
U= 3.9375 j
To find the spring potential energy, we can use the formula:
Potential Energy (PE) = (1/2) * k * x^2
where k is the spring constant and x is the displacement from the equilibrium position.
In this case, the spring is stretched by 15 cm, which is equivalent to 0.15 m, and the spring constant is 350 N/m. Let's substitute these values into the formula:
PE = (1/2) * 350 N/m * (0.15 m)^2
Calculating this expression will give us the spring potential energy.
To find the force required for this stretch, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from the equilibrium position.
Hooke's Law formula:
Force (F) = k * x
where F is the force applied, k is the spring constant, and x is the displacement.
In this case, the spring constant is given as 350 N/m and the displacement is 0.15 m. Let's substitute these values into the formula:
F = 350 N/m * 0.15 m
Calculating this expression will give us the force required for this stretch.