Question

A 1.0 kg mass that can move along the x-axis experiences the potential energy u=(x²-x)j, where x is in m. The mass has velocity vx = 2.0 m/s at position x = 1.0 m. At what position has the velocity slowed to 1.0 m/s?

Answer 1

To find the position where the velocity of the 1.0 kg mass has slowed to 1.0 m/s, you can equate the kinetic energy to the change in potential energy. By rearranging the equation and using the quadratic formula, you can find the two possible positions where the velocity has slowed to 1.0 m/s: x = 0.79 m and x = 1.21 m.

Step-by-step explanation:

To find the position where the velocity of the 1.0 kg mass has slowed to 1.0 m/s, we need to equate the kinetic energy (KE) to the change in potential energy (PE) from the initial position to the desired position. Since the kinetic energy is given by KE = 0.5 * m * v^2 and the potential energy is given by PE = x^2 - x, we can set up the equation:

0.5 * 1.0 * (1.0)^2 = (desired position)^2 - (desired position)

Simplifying this equation, we get:

0.5 = (desired position)^2 - (desired position)

By rearranging the equation, we can express it as a quadratic equation:

(desired position)^2 - (desired position) - 0.5 = 0

Using the quadratic formula, we can find the two possible positions:

x = (-b ± √(b^2 - 4ac)) / (2a)

Substituting the values a = 1, b = -1, and c = -0.5 into the formula, we find that the two possible positions where the velocity has slowed to 1.0 m/s are approximately x = 0.79 m and x = 1.21 m.