A 60 point 0 kilograms runner is moving at 6.00 meters per second and speeds up to 9.00 meters per second The runner’s change in kinetic energy is 1 comma 350 cap j How much work was done?

By the work–energy theorem, the net work done on the runner equals the change in kinetic energy: W = ΔKE. Since the runner’s kinetic energy increases by 1,350 J, the work done is positive 1,350 J. The correct choice is D, 1,350 J.

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