Maxx Crosby was dominant on Sunday, earning the AFC Defensive Player of the Week award for Week 2 of the 2024 NFL season.
His all-around performance played a pivotal role in helping the Las Vegas Raiders secure a crucial road win against the Baltimore Ravens. Crosby’s relentless play was a key factor in the team’s bounce-back victory, evening their record at 1-1 for the season.
Crosby didn’t miss a single snap during the game, finishing with six tackles (five solo), 2.0 sacks, four tackles for loss, two quarterback hits, and a pass deflection. His individual efforts stood out, leading all defensive linemen in solo tackles and registering the most tackles for loss among AFC players in Week 2.
This is the fifth time Crosby has received the AFC Defensive Player of the Week honor and the second time he’s won it after a standout game against the Ravens. His performance solidifies his status as one of the NFL’s elite pass rushers. With this latest accolade, Crosby holds the most Defensive Player of the Week honors in Raiders’ franchise history.
As the season progresses, he remains a frontrunner for NFL Defensive Player of the Year, but his focus will be on leading the defense in their upcoming home opener against the Carolina Panthers.
