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James Conner wins AFC Offensive Player of the Week honors for game against Browns

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The Steelers running back won the first player of the week award of his career on Wednesday for his performance against the Cleveland Browns in Week 8.

NFL: Cleveland Browns at Pittsburgh Steelers Jeffrey Becker-USA TODAY Sports

A Pittsburgh Steelers player winning player of the week honors has become something of a biweekly affair in 2018 after James Conner was named AFC Offensive Player of the Week on Wednesday for his efforts against the Cleveland Browns in Week 8.

Conner rushed for 146 yards and 2 touchdowns from 24 carries on the day and added another 66 yards from 5 receptions. Not only was it his highest single game rushing total of the season, it was also his second best ever receiving total and the first time he recorded more than 200 scrimmage yards in a single game in his career.

Having scored two touchdowns in each of his last three games, Conner has set a new record in Pittsburgh, achieving a feat the great Steelers running backs like Franco Harris and Jerome Bettis never managed to do.

The Pittsburgh running back becomes the third Steelers player to win a weekly award in 2018. T.J. Watt has won defensive player of weeks honors twice this season, once in Week 1 and again in Week 5, while Ben Roethlisberger won the offensive award in Week 3. Much like they are for Watt, the Browns must already be Conner’s favorite team to play.

Currently third in the NFL in rushing with 599-yards, Conner only sits behind Todd Gurley of the Los Angeles Rams (800) and Ezekiel Elliott of the Dallas Cowboys (619). He is also second in the league with nine rushing touchdowns behind Gurley (11). If he can maintain this pace throughout the remainder of the season, Conner would eclipse Le’Veon Bell’s career best rushing total with the Steelers of 1,361 yards.