a horse pulls the wagon of 3075 kg from rest against constant resistance of 90 N. the pull exerted initially is 600 N and it decreases uniformly with the distance covered to 400N at a distance of 15m from the start. find the velocity at this time.

pls answer

Since force varies uniformly Hence, average force=(600+400)/2= 500 N K.E = 500×15 - 90×15 =6150 N Now v^2 = (2×6150)÷3075 V = 2 m/s

Since force varies uniformly Hence, average force=(600+400)/2= 500 N K.E = 500×15 - 90×15 =6150 N Now v^2 = (2×6150)÷3075 V = 2 m/s

Force is variable so average force = (Final force+initial force)/2=(600+400)/2=500 N
resistance ia constt so net force acting=500-90=410 N
work done =F.x=410.15=6150J = Change in KE = KEf-KEi
KEi=0(given, as object at rest initially)
so (1/2)(mass)(v^2)=6150
mass=3075
hence
v^2=6150*2/3075=4
so v=2

Someome please tell why did we calculated (400+600)/2

Someone please tell why did we calculated avergae force...please tell

Hope u get it

Hope you get it