A 325-kg boat is sailing 15.0° north of east at a speed of 2.00 m/s. Thirty seconds later, it is sailing 35.0° north of east at a speed of 4.00 m/s. During this time, three forces act on the boat: a 31.0-N force directed 15.0° north of east (due to an auxiliary engine), a 23.0-N force directed 15.0° south of west (resistance due to the water), and (due to the wind). Find the magnitude and direction of the force . Express the direction as an angle with respect to due east.

1 answer

Fnet=Fb+Fr+Fw
Fnet=ma
a=change v/change t = 2/30=0.0667
fnet=325(0.0667)=21.67N

21.67=31+(-23)+Fw
Fw=13.7N 15 North of East

35.0 is an irrelivant detail - fw can be calculated with the given times and masses