Solving for Velocity: 8.2 kg Object Moving Along x-Axis

[cjwiegmann]

Homework Statement

An 8.2 kg object is moving in the positive direction of an x axis. When it passes through x = 0, a constant force directed along the axis begins to act on it. The figure below gives its kinetic energy K versus position x as it moves from x = 0 to x= 5.0 m; K0 = 27.5 J. The force continues to act. What is v when the object moves back through x = -3.0 m?

Homework Equations

The Attempt at a Solution

 

[member 553083]

When the object passes through x = 0, its kinetic energy is 27.5 J. We can use the equation K = 1/2mv^2 to solve for v at x = 0: 27.5 J = (1/2)(8.2 kg)(v^2)v = sqrt(54.5 J/8.2 kg)v = 3.6 m/sSince the force continues to act on the object as it moves in the negative direction, the object's kinetic energy will increase. When the object passes through x = -3.0 m, we can again use the equation K = 1/2mv^2: K = 27.5 J + mv^2 K = 27.5 J + (8.2 kg)(v^2)27.5 J + (8.2 kg)(v^2) = K (8.2 kg)(v^2) = K - 27.5 J v = sqrt((K - 27.5 J)/(8.2 kg)) We know that K = 55 J when the object is at x = 5.0 m, so: v = sqrt((55 J - 27.5 J)/(8.2 kg)) v = sqrt(27.5 J/8.2 kg) v = 4.3 m/s