What is its velocity at x=1m, 2m, and 3m?

The area under the graph gives the work done by the force (Integral[F.dx]). This gives the increase in kinetic energy, KE)Initial KE = mv^2/2 = 0.6x2^2/2 = 1.2JAt 1m, Work total done = area up to 1m = 1 x (10+15)/2 = 12.5JNew total KE = 1.2 + 12.5 = 13.7J From KE = mv^2/2, v = sqrt(2KE/m) = sqrt(2x13.7/0.6) = 6.8m/sAt 2m, Work total done = area up to 2m = 2 x (15+ 5)/2 = 20JNew total KE = 1.2 + 20 = 21.2Jv = sqrt(2KE/m) = sqrt(2x21.2/0.6) = 8.4m/sAt 3m - you can do it yourself, to make sure you understand the method. I get 8.9m/s... Show More

The "area" under a curve that has force as the y-axis and distance as the x-axis has units of energy.The energy involved in this problem is kinetic: (1/2)mv^2First, compute the original kinetic energy:(1/2) m v^2 = (1/2) (0.6 kg)(2^2) = 1.2 Joules of kinetic energy at x=0 (KE0)The "area" under the curve from x = 0 to x = 1m is 12.5 Newton*meters or 12.5 JoulesIF the direction of force is the same as the direction of motion, then we just add the energies together to get a new kinetic energy:1.2 Joules + 12.5 Joules = 13.7 Joules of kinetic energy at x=1m (KE1)The "area" under the curve from x = 1m to x = 2m is 7.5 Newton*meters or 7.5 Joulesso the kinetic energy is raised to:13.7 Joules + 7.5 Joules = 21.2 Joules of kinetic energy at x=2m (KE2)The "area" under the curve from x = 2m to x = 3m is 2.5 Newton*meters or 2.5 Joulesso the kinetic energy is raised to:21.2 Joules + 2.5 Joules = 23.7 Joules of kinetic energy at x=3m (KE3)To convert these values of kinetic energy into speed, use:KE = (1/2) m v^2solving for v:v = square root of (2*KE / m) (Equation A)Since m = 0.6kg:v = square root of (2*KE / 0.6kg) (Equation B)Just plug in the values of KE1, KE2 and KE3 into Equation B to get the speeds: They should be 6.76 m/s, 8.41 m/s, and 8.89 m/sPlease double check my calculations... Show More