Suppose a 0.74-kg mass on a spring that has been compressed 0.100 m has elastic potential energy of 1.20 J. What is the spring constant?
Do I use 0.100m as x in the equation PE=1/2 kx^2??
Thats all I am confused about.
Thanks!
Nevermind I was correct, no need to answer this question.
To find the spring constant, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. Mathematically, it can be expressed as:
F = -kx
Where F is the force exerted by the spring, k is the spring constant, and x is the displacement of the spring from its equilibrium position.
In this problem, we are given the mass of the object (0.74 kg), the displacement of the spring (0.100 m), and the elastic potential energy (1.20 J). We can use this information to find the spring constant.
We know that the elastic potential energy is given by:
PE = (1/2)kx^2
Where PE is the elastic potential energy, k is the spring constant, and x is the displacement of the spring.
Substituting the given values into the equation, we have:
1.20 J = (1/2)k(0.100 m)^2
Simplifying the equation:
1.20 J = (1/2)k(0.0100 m^2)
Now, let's solve for k.
Multiply both sides of the equation by 2:
2 * 1.20 J = k(0.0100 m^2)
2.40 J = k(0.0100 m^2)
Divide both sides of the equation by 0.0100 m^2:
2.40 J / (0.0100 m^2) = k
Simplifying the equation:
240 N/m = k
Therefore, the spring constant is 240 N/m.