Rob Gronkowski Wins AFC Offensive Player Of The Week
For the second time in his career, New England Patriots tight end Rob Gronkowski has been named the AFC’s Offensive Player of the Week for Week 15 of the 2017 season.
Gronk was recognized for his performance in this past Sunday’s game against the Pittsburgh Steelers, where he finished with nine catches totaling a career high of 168 yards in one game.
Career-high 168 yards + critical 2-point conversion = AFC Offensive Player of the Week: https://t.co/3FPN9SwpbY pic.twitter.com/gUI7DPwJqb
— New England Patriots (@Patriots) December 20, 2017
During the game-winning drive that sent the Patriots to a 27-24 win over the Steelers, Gronkowski picked up three receptions, totaling 69 yards.
Gronkowski’s performance on Sunday brings him to 1,017 total yards for 2017, which puts him over 1,000 yards for the fourth time in his career.
