Aforce of 70n is applied to a 7kg block that was at rest. the surface is flat and frictionless. how far did the block travel in 10s

You can use newtons second law to solve this F=ma where a=dv/dt so this gives F=m(dv/dt) where dv/dt is the change velocity with respect to time. since the object started from rest we can say that initial velocity=0 and we can assume the time started at t=0 so the change in velocity = final velocity - 0 = final velocity we'll just call it v. and for the change in time we have 10s-0s = 10s so t=10s. this leads to the equation F=m(v/t) so rearranging to solve for velocity we have (F*t)/m=v so plugging in numbers we have F=70N, t=10s and m=7kg so we have (70kg*m/s^2*10s)/7kg=v so v = 100m/s this looks pretty good since all the units canceled correctly and left us with the proper units for velocity. now with the final velocity we can solve for the distance traveled using the kinematic equation d=(t(Vf+Vi)/2) so again plugging in numbers we get d= (10s(100m/s+0))/2= 500m.

Answer choices? if there is any

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