Twisting the knife just a little bit more.
Saquon Barkley is NFC Offensive Player of the Week for Week 7 of the 2024 NFL season, according to an official announcement.
No surprise here.
Barkley was outstanding in his return to MetLife Stadium for a revenge game against the New York Giants. The star running back had 17 carries for 176 rushing yards (10.4 average) and one touchdown in addition to two catches for 11 receiving yards.
Had the Eagles not pulled their starters early in the fourth quarter due to having a big lead, Barkley easily could’ve surpassed his previous single-game career high of 189 yards.
Barkley previously won NFC Offensive Player of the Week for his three-touchdown performance against the Green Bay Packers in Week 1.
This honor marks the fifth time in his career that Barkley has been named NFC POTW. And it’s the first time he’s ever won it twice in the same season.
With Barkley looking unstoppable and plenty of games left to play, even more awards could be on the way.
In related news, Giants owner John Mara is not sleeping well these days.
