A 15.0-kilogram mass is moving at 7.50 meters per second on a horizontal, frictionless surface. What is the
total work that must be done on the mass to increase its speed to 11.5 meters per second?
(1) 120. J (3) 570. J
(2) 422 J (4) 992 J
The answer is 570 J. The kinetic energy has the formula of 1/2mV². The total work in this process W= 1/2m(V2²-V1²) = 1/2 * 15.0 * (11.5²-7.50²) = 570 J.
