2.50 kg particle (the function U(x) has the form bx2 and the vertical axis scale is set by Us = 2.0 J).(a) If the particle passes through the equilibrium position with a velocity of 55.0 cm/s, will it be turned back before it reaches x = 18.0 cm?(b) If yes, at what position (in cm), and if no, what is the speed of the particle (in cm/s) at x = 18.0 cm?
I don't know what you mean by Us = 2.0 J
This looks like a mass on a spring problem for which conservation of energy should be used. The sum of potential and kinetic energy equals a constant. At the equilibrium position, the kinetic energy is
(1/2)M V^2 = (1/2)(2.5)(0.55)^ = 2
= 0.378 J
Yhat equals the consnta total energy of the system.
You need to determine the spring constant somehow so you can calculate the potential energy when x = 0.17 m. From that, you can get the speed.
To determine whether the particle will be turned back before reaching x = 18.0 cm, we need to calculate the potential energy at the equilibrium position and the potential energy at x = 18.0 cm.
First, let's find the potential energy at the equilibrium position. The potential energy function U(x) is given by U(x) = bx^2, and the vertical axis scale is set by Us = 2.0 J. Since U(x) has the form bx^2, it implies that b = Us/x^2.
Given Us = 2.0 J, and the equilibrium position x = 0, we can substitute these values into the equation to find b:
b = Us/x^2 = 2.0 J/ (0 cm)^2 = ∞ J/cm^2
Now, let's find the potential energy at x = 18.0 cm:
U(18.0 cm) = b(18.0 cm)^2 = (∞ J/cm^2)(18.0 cm)^2 = ∞ J
Since the potential energy at x = 18.0 cm is infinite (∞ J), the particle cannot reach that position. Therefore, the answer to part (a) is NO, the particle will not be turned back before reaching x = 18.0 cm.
To answer part (b), since the particle cannot reach x = 18.0 cm, we do not need to calculate its speed.