A highway goes up a hill, rising at a constant rate of 1.00 m for every 32 m alo

Question

A highway goes up a hill, rising at a constant rate of 1.00 m for every 32 m along the road. A truck climbs this hill at constant speed vup = 17 m/s, against a resisting force (friction) f equal to 1/16 of the weight of the truck. Now the truck comes down the same hill, using the same power as it did going up. Find vdown, the constant speed with which the truck comes down the hill.
ASSUME: the resisting force (friction) has the same magnitude going up as going down.

?m/s

<<PLEASE EXPLAIN ME STEP BY STEP>>

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Explanation / Answer

power initially=(w*sin theta+friction)*v=(w*1/32+w/16)*17=(3/32)*17*w force downwards=(w*sin theta-friction)*v=(-w*1/32+w/16)=1/32 w as force is one tird vwill be 3 times so v=51

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