Question

A single conservative force acts on a 5.00 kg particle. The equation Fx = (2x + 4) N describes this force, where x is in meters. As the particle moves along the x axis from x = 3.00 m to x = 7.00 m, calculate the following. (a) the work done by this force on the particle J (b) the change in the potential energy of the system J (c) the kinetic energy the particle has at x = 7.00 m if its speed is 3.00 m/s at x = 3.00 m

Answer 1

a) At x = 6.2 m, F(6.2) = 2*6.2 + 4 = 16.4 N At x = 2.2 m, F(2.2) = 2*2.2 + 4 = 8.4 N Resultant F = 16.4 - 8.4 = 8 N Work done, W = resultant force, F * distance, d = 8 N * (6.2 - 2.2) m = 32 Nm = 32 J b) depends on any change in height (vertical direction) of particle. if x is the horizontal direction (?) w.r.t. the direction of gravity force, then Potential energy (grav.) U = m*g*h = 5 * 9.81 * 0 = 0 J i.e. no change in potential energy c) Initial kinetic energy = 0.5*m*v^2 = 0.5*5*3^2 = 22.5 J total energy supplied = final kinetic energy(KE2) - initial kinetic energy => 32 = KE2 - 22.5 => KE2 = 32+22.5 = 54.5 J