Philadelphia Eagles running back Saquon Barkley won the 2024 Offensive Player of the Year award Thursday.
Cincinnati Bengals quarterback Joe Burrow, Bengals wide receiver Ja'Marr Chase, Baltimore Ravens running back Derrick Henry, and Ravens signal-caller Lamar Jackson were the other finalists for the award.
Barkley claimed 406 points and 35 first-place votes, according to Rob Maaddi of The Associated Press. He was followed by Jackson (183 points) and Chase (171).
The Penn State product excelled in his first season with the Eagles. He totaled 2,005 rushing yards, 278 receiving yards, and 15 total touchdowns. His rushing and scrimmage yard marks both led the NFL even though he sat out the regular-season finale with Philadelphia's playoff seeding locked in.
Barkley, who turns 28 on Feb. 9, is just the eighth player to reach 2,000 rushing yards in a single campaign. He earned Pro Bowl and first-team All-Pro honors for his efforts this season.
The 2018 first-rounder is the fourth running back to win Offensive Player of the Year since 2017. Former Los Angeles Rams runner Todd Gurley won the award in 2017, Henry took home the honor in 2020, and 49ers dual-threat Christian McCaffrey was last year's winner.
