Rob Gronkowski Named AFC Offensive Player of the Week
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For the second time in his career, New England Patriots tight end Rob Gronkowski has been named AFC Offensive Player of the Week for his performance in the Patriots 27-24 victory at Pittsburgh on Sunday. It becomes the second time Gronkowski has earned a weekly honor and the first since Week 14 of the 2011 season.
Per the Patriots offical press release, “Gronkowski finished the game with nine receptions for a career-high 168 yards. The 168-yards pushed his 2017 total to 1,017 yards, giving him his fourth 1,000-yard season. Gronkowski joins Tony Gonzalez and Jason Witten as the only NFL tight ends with four 1,000-yard seasons. In addition, to his 168-yard performance, Gronkowski converted a key 2-point play after the game-winning touchdown to give the Patriots a 27-24 lead. During the game-winning drive, Gronkowski caught three passes for 69 yards with back-to-back 26-yard receptions followed by a 17-yard catch. He had four receptions of at least 20 yards in the game.”
https://twitter.com/patriots/status/942580537473900544
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The honor marks the sixth time in 2017 a Patriot has recieved a weekly award. Quarterback Tom Brady has earned three AFC Offensive Player of the Week honors during Week 2, 3, and 10 and AFC Offensive Player of the Month for November. Running back Dion Lewis earned an AFC Special Teams Player of the Week in Week 10. While kicker Stephen Gostkowski was named AFC Special Teams Player of the Week for Week.
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